An electric dipole is placed in an external uniform electric field. The net electric force on the dipole
(A) is always zero
(B) depends on the orientation of the dipole
(C) can never be zero
(D) depends on the dipole moment of the dipole
Correct Answer: (A) is always zero
View SolutionStep 1: Understanding the Question:
An electric dipole has two equal and opposite charges, +q and -q, a small distance apart. We are asked about the net force on it when the external electric field is the same everywhere (uniform).
Step 2: Key Formula or Approach:
The force on a charge q in a field \(\vec{E}\) is \(\vec{F} = q\vec{E}\). The net force is the vector sum of the forces on the two charges.
Step 3: Force on Each Charge:
The force on +q is \(+q\vec{E}\). The force on -q is \(-q\vec{E}\). The field is the same at both places, because it is uniform.
Step 4: Net Force:
\[ \vec{F}_{net} = q\vec{E} + (-q)\vec{E} = 0 \] The two forces are equal in size and opposite in direction, so they cancel. This holds for every orientation of the dipole. They act at different points, so they can form a torque, but the net force stays zero.
Step 5: Checking Each Option:
Option 1 says the net force is always zero. This is true for a uniform field. Option 2 is wrong because orientation changes the torque, not the net force. Option 3 is wrong because the net force does become zero. Option 4 is wrong because the dipole moment \(p = q \cdot 2a\) affects the torque \(\tau = pE\sin\theta\), not the net force.
Final Answer:
The net force on a dipole in a uniform electric field is always zero, so option 1 is correct.
\[ \boxed{1} \]
Question 2:
Identify the correct statements from the following : A. The charge on a body can have any value greater than the charge on an electron or proton. B. Gauss's theorem is valid for a closed surface of any shape and for any general charge distribution. C. The net flux through a closed surface due to a charge lying outside the closed surface is zero. D. Gauss's theorem is applicable to any field which obeys the inverse square law Choose the correct answer from the options given below:
(A) A, B and C only
(B) A, B and D only
(C) A, B, C and D
(D) B, C and D only
Correct Answer: (D) B, C and D only
View SolutionStep 1: Understanding the Question:
We must decide which of the four statements about charge and Gauss's theorem are true, and then pick the matching option.
Step 2: Statement A:
Charge is quantised. Any charge is an integer multiple of the electron charge: \(q = ne\). So a body cannot have any value. Only whole-number multiples of \(e\) are possible. Statement A is false.
Step 3: Statement B:
Gauss's theorem says \(\oint \vec{E}\cdot d\vec{S} = q_{enc}/\varepsilon_0\). It holds for a closed surface of any shape and for any charge distribution. Statement B is true.
Step 4: Statement C:
Field lines from an outside charge enter the closed surface and leave it. The inward flux equals the outward flux, so the net flux is zero. Statement C is true.
Step 5: Statement D:
Gauss's theorem depends on the field falling as \(1/r^2\). The area of a sphere grows as \(r^2\), so the flux stays constant. Any field that obeys the inverse square law follows the theorem. Statement D is true.
Step 6: Matching the Options:
True statements are B, C and D. Option 1 includes A, so it is wrong. Option 2 includes A and leaves out C, so it is wrong. Option 3 includes A, so it is wrong. Option 4 is B, C and D only.
Final Answer:
Statements B, C and D are correct, so option 4 is the answer.
\[ \boxed{\text{B, C and D only}} \]
Question 3:
Match the LIST-I with LIST-II
LIST-I
LIST-II
A. \(E\) is independent of \(r\)
I. For a point charge
B. \(E\) is inversely proportional to \(r\)
II. For a short dipole
C. \(E\) is inversely proportional to \(r^2\)
III. For a uniform line charge
D. \(E\) is inversely proportional to \(r^3\)
IV. For a uniformly charged infinite plane sheet
Choose the correct answer from the options given below:
(A) A-I, B-II, C-III, D-IV
(B) A-IV, B-III, C-I, D-II
(C) A-IV, B-III, C-II, D-I
(D) A-III, B-IV, C-I, D-II
Correct Answer: (B) A-IV, B-III, C-I, D-II
View SolutionStep 1: Understanding the Question:
We must match each way the field \(E\) changes with distance \(r\) to the charge arrangement that gives it.
Step 2: Infinite Plane Sheet:
For a uniformly charged infinite sheet, \(E = \sigma/2\varepsilon_0\). It has no \(r\) in it, so \(E\) is independent of distance. So A matches IV.
Step 3: Uniform Line Charge:
For a long line charge, \(E = \lambda/2\pi\varepsilon_0 r\). So \(E \propto 1/r\). B matches III.
Step 4: Point Charge:
For a point charge, \(E = kq/r^2\). So \(E \propto 1/r^2\). C matches I.
Step 5: Short Dipole:
On the axis or equator of a short dipole, \(E \propto p/r^3\). So D matches II.
Step 6: Checking the Options:
We have A-IV, B-III, C-I, D-II. Option 1 has A-I, which is wrong. Option 3 has C-II, which is wrong. Option 4 has A-III, which is wrong. Option 2 matches exactly.
Final Answer:
The correct matching is A-IV, B-III, C-I, D-II, which is option 2.
\[ \boxed{\text{A-IV, B-III, C-I, D-II}} \]
Question 4:
A charge q is placed at the center of a line joining two equal positive charges Q. The system of the three charges will be in equilibrium, if q is equal to
(A) \(-\frac{Q}{2}\)
(B) \(-\frac{Q}{4}\)
(C) \(+\frac{Q}{4}\)
(D) \(+\frac{Q}{2}\)
Correct Answer: (B) \(-\frac{Q}{4}\)
View SolutionStep 1: Understanding the Question:
Two charges Q sit a distance \(d\) apart. A charge q sits at the midpoint. All three charges must be in equilibrium, so the net force on each one must be zero.
Step 2: Key Formula or Approach:
Use Coulomb's law \(F = kq_1q_2/r^2\). It is enough to look at one of the end charges. The middle charge is in equilibrium by symmetry, because the two Q charges push or pull it equally in opposite directions.
Step 3: Force on One Charge Q:
The other Q repels it with \[ F_1 = \frac{kQ^2}{d^2} \] The middle charge is at \(d/2\), so it acts with \[ F_2 = \frac{kQ|q|}{(d/2)^2} = \frac{4kQ|q|}{d^2} \]
Step 4: Sign of q:
The force from the other Q pushes the charge away from the centre. To cancel it, the middle charge must pull toward the centre. So q must be opposite in sign to Q. That means q is negative.
Step 5: Balancing the Forces:
\[ \frac{4kQ|q|}{d^2} = \frac{kQ^2}{d^2} \] \[ |q| = \frac{Q}{4} \] With the negative sign, \(q = -Q/4\).
Step 6: Checking the Options:
Options 3 and 4 are positive, so they cannot balance the repulsion. Option 1 gives \(-Q/2\), which pulls too strongly. Option 2, \(-Q/4\), balances the forces exactly.
Final Answer:
The middle charge must be \(q = -Q/4\), which is option 2.
\[ \boxed{q=-\frac{Q}{4}} \]
Question 5:
A metal wire is bent in the shape of a circle of 10 cm radius. It is given a charge of 200 \(\mu\)C which spreads on it uniformly. The electric potential at its center is
(A) \(3 \times 10^{6}\) V
(B) \(6 \times 10^{6}\) V
(C) \(9 \times 10^{6}\) V
(D) \(18 \times 10^{6}\) V
Correct Answer: (D) \(18 \times 10^{6}\) V
View SolutionStep 1: Understanding the Question:
A uniformly charged ring has total charge \(Q = 200\ \mu C\) and radius \(R = 10\) cm. We need the potential at the centre.
Step 2: Key Formula or Approach:
Every small piece of charge is at the same distance R from the centre. So the potential adds up simply: \[ V = \frac{kQ}{R} \]
Step 3: Convert the Units:
\(Q = 200\times10^{-6} = 2\times10^{-4}\) C. \(R = 10\) cm \(= 0.1\) m. \(k = 9\times10^{9}\) N m2/C2.
Step 5: Checking the Options:
Options 1, 2 and 3 (\(3\), \(6\) and \(9\) times \(10^6\) V) are smaller than the value we found. They come from dropping a factor in the calculation. Option 4 matches.
Final Answer:
The potential at the centre is \(18\times10^{6}\) V, so option 4 is correct.
\[ \boxed{18\times10^{6}\ \text{V}} \]
Question 6:
Three capacitors of equal capacitance, when connected in series have net capacitance \(C_1\) and when connected in parallel have net capacitance \(C_2\). The ratio of \(C_1\) and \(C_2\) is
(A) \(\frac{1}{3}\)
(B) 3
(C) \(\frac{1}{9}\)
(D) 9
Correct Answer: (C) \(\frac{1}{9}\)
View SolutionStep 1: Understanding the Question:
Three identical capacitors, each of capacitance C, are joined in series to give \(C_1\) and in parallel to give \(C_2\). We need \(C_1 : C_2\).
Step 2: Series Combination:
In series, the reciprocals add: \[ \frac{1}{C_1} = \frac{1}{C}+\frac{1}{C}+\frac{1}{C} = \frac{3}{C} \] So \(C_1 = C/3\).
Step 3: Parallel Combination:
In parallel, the capacitances add: \[ C_2 = C + C + C = 3C \]
Step 4: Take the Ratio:
\[ \frac{C_1}{C_2} = \frac{C/3}{3C} = \frac{1}{9} \]
Step 5: Checking the Options:
Option 1 (1/3) would come from forgetting the factor 3 in the parallel value. Options 2 and 4 are larger than 1, but the series value is smaller than the parallel one. Option 3, 1/9, is right.
Final Answer:
The ratio \(C_1 : C_2\) is 1/9, which is option 3.
\[ \boxed{\frac{1}{9}} \]
Question 7:
A 800 pF capacitor is charged by a 100 V battery. After some time, the battery is disconnected. The capacitor is then connected across another 800 pF capacitor. The electrostatic energy stored in the combination will be
(A) \(2 \times 10^{-6}\) J
(B) \(3.8 \times 10^{-6}\) J
(C) \(4.2 \times 10^{-6}\) J
(D) \(6 \times 10^{6}\) J
Correct Answer: (A) \(2 \times 10^{-6}\) J
View SolutionStep 1: Understanding the Question:
A 800 pF capacitor is charged to 100 V and then cut off from the battery. It is then joined to an identical uncharged 800 pF capacitor. We need the total energy afterwards.
Step 2: Charge Stored at the Start:
\[ Q = CV = 800\times10^{-12}\times100 = 8\times10^{-8}\ \text{C} \] Since the battery is disconnected, this total charge stays the same.
Step 3: Energy Stored at the Start:
\[ U_i = \frac{1}{2}CV^2 = \frac{1}{2}\times800\times10^{-12}\times(100)^2 = 4\times10^{-6}\ \text{J} \]
Step 4: After Connecting:
The two capacitors are joined in parallel, so the common capacitance is \(C' = 1600\) pF. The charge is still \(Q\). The common voltage is \[ V' = \frac{Q}{C'} = \frac{8\times10^{-8}}{1600\times10^{-12}} = 50\ \text{V} \]
Step 5: Final Energy:
\[ U_f = \frac{1}{2}C'V'^2 = \frac{1}{2}\times1600\times10^{-12}\times(50)^2 = 2\times10^{-6}\ \text{J} \] Half of the energy is lost, mostly as heat and radiation while the charge flows.
Step 6: Checking the Options:
Option 1 gives \(2\times10^{-6}\) J and matches. Options 2 and 3 are not obtained by any correct step. Option 4 has a positive power, \(10^6\), which is far too large.
Final Answer:
The final energy is \(2\times10^{-6}\) J, so option 1 is correct.
\[ \boxed{2\times10^{-6}\ \text{J}} \]
Question 8:
Match the LIST-I with LIST-II
LIST-I (Quantity)
LIST-II (Formula)
A. Electromotive force
I. Drift velocity / Electric field
B. Electric field
II. Current / Area
C. Current density
III. Electric force / Charge
D. Charge mobility
IV. Work / Charge
Choose the correct answer from the options given below:
(A) A-I, B-II, C-III, D-IV
(B) A-IV, B-III, C-II, D-I
(C) A-IV, B-III, C-I, D-II
(D) A-III, B-IV, C-II, D-I
Correct Answer: (B) A-IV, B-III, C-II, D-I
View SolutionStep 1: Understanding the Question:
We must pair each physical quantity in List-I with the formula in List-II that defines it. All four quantities come from current electricity.
Step 2: Match A, electromotive force:
EMF is the work done per unit charge by the source in moving charge around the circuit. So EMF = Work / Charge, which is item IV. A matches IV.
Step 3: Match B, electric field:
Electric field is the force felt by a unit test charge. So E = Force / Charge, which is item III. B matches III.
Step 4: Match C, current density:
Current density is the current flowing through unit area, \(J = I/A\). This is item II. C matches II.
Step 5: Match D, charge mobility:
Mobility is the drift velocity gained per unit electric field, \(\mu = v_d/E\). This is item I. D matches I.
Step 6: Compare with the options:
The full matching is A-IV, B-III, C-II, D-I. Option 1 wrongly pairs A with I. Option 3 swaps C and D. Option 4 pairs A with III. Only option 2 is fully right.
Final Answer:
The matching is A-IV, B-III, C-II, D-I, which is option 2.
\[ \boxed{\text{Option 2: A-IV, B-III, C-II, D-I}} \]
Question 9:
Arrange the following materials in the ascending order of their electrical conductivities. A. Rubber B. Iron C. Carbon D. Copper Choose the correct answer from the options given below:
(A) A, B, C, D
(B) D, B, C, A
(C) A, C, B, D
(D) C, A, B, D
Correct Answer: (C) A, C, B, D
View SolutionStep 1: Understanding the Question:
We need to rank four materials from the poorest to the best conductor of electricity.
Step 2: Rubber:
Rubber is an insulator. Its conductivity is about \(10^{-13}\) S/m or even lower, so it is the lowest of the four.
Step 3: Carbon:
Carbon (graphite) conducts poorly compared with metals. Its conductivity is about \(10^{4}\) S/m, which is much above rubber but far below metals.
Step 4: Iron:
Iron is a metal with conductivity of about \(10^{7}\) S/m. It is higher than carbon.
Step 5: Copper:
Copper has one of the highest conductivities among common metals, about \(6\times10^{7}\) S/m. It is higher than iron.
Step 6: Arrange in ascending order:
The order is Rubber < Carbon < Iron < Copper, that is A, C, B, D. This is option 3. Option 1 puts iron before carbon. Option 2 is the descending-type order. Option 4 puts carbon before rubber.
Final Answer:
The ascending order of conductivity is A, C, B, D, which is option 3.
\[ \boxed{\text{Option 3: A, C, B, D}} \]
Question 10:
Identify the correct statements among the following: A. A current carrying wire is electrically charged. B. The drift velocity of electrons in a metallic wire will decrease if the temperature of the wire is increased. C. The emf of a cell depends on the internal resistance of the cell. D. The internal resistance of dry cells is much higher than the common electrolytic cells. Choose the correct answer from the options given below:
(A) A, B and D only
(B) A and D only
(C) B and D only
(D) A, C and D only
Correct Answer: (C) B and D only
View SolutionStep 1: Understanding the Question:
We must decide which of the four statements are true, then pick the option that lists exactly those.
Step 2: Check statement A:
A current carrying wire has as many positive charges (lattice ions) as free electrons. So it is electrically neutral overall. A is FALSE.
Step 3: Check statement B:
When temperature rises, the ions vibrate more and collisions with electrons become more frequent. The relaxation time falls. Drift velocity \(v_d = eE\tau/m\) then decreases. B is TRUE.
Step 4: Check statement C:
The emf of a cell is the potential difference across its terminals when no current flows. It depends on the chemicals and electrodes, not on the internal resistance. C is FALSE.
Step 5: Check statement D:
Dry cells have a thick paste as electrolyte and small electrode area, so their internal resistance is much higher than that of ordinary liquid electrolytic cells. D is TRUE.
Step 6: Pick the option:
Only B and D are true. That is option 3. Options 1, 2 and 4 include A or C, which are false.
Final Answer:
Only statements B and D are correct, so option 3.
\[ \boxed{\text{Option 3: B and D only}} \]
Question 11:
At what temperature the resistance of a conductor becomes 20% more than its resistance at \(27^\circ\) C? (The value of the temperature coefficient of resistance of the conductor is \(2.0 \times 10^{-4}\)/K.)
(A) 900 K
(B) 1100 K
(C) 1300 K
(D) 1450 K
Correct Answer: (C) 1300 K
View SolutionStep 1: Understanding the Concept:
Resistance of a conductor changes almost linearly with temperature. The temperature coefficient \(\alpha\) tells the fractional change per kelvin.
Step 2: Key Formula:
\[ R_T = R_{27}\,[1 + \alpha (T - T_0)] \] where \(T_0 = 27 ^\circ C = 300\) K.
Step 3: Put the values:
We need \(R_T = 1.2\,R_{27}\). So \[ 1.2 = 1 + 2.0\times10^{-4}\,\Delta T \] \[ 2.0\times10^{-4}\,\Delta T = 0.2 \] \[ \Delta T = \frac{0.2}{2.0\times10^{-4}} = 1000 \text{ K} \]
Step 4: Find the final temperature:
\(T = 27 ^\circ C + 1000 = 1027 ^\circ C\). In kelvin, \(T = 300 + 1000 = 1300\) K.
Step 5: Check the options:
900 K and 1100 K give a rise of only 600 K and 800 K, so the resistance would rise only 12% and 16%. 1450 K gives a rise of 1150 K, that is 23%. Only 1300 K gives exactly 20%.
Final Answer:
The conductor reaches 20% more resistance at 1300 K.
\[ \boxed{1300\ \text{K}} \]
Question 12:
In a Wheatstone bridge arrangement (P, Q, R and S), the resistors P and Q are nearly equal. The bridge is balanced when R = 500 \(\Omega\). On interchanging P and Q, the value of R for balancing is 505 \(\Omega\). The value of S will be:
(A) 500 \(\Omega\)
(B) 502.5 \(\Omega\)
(C) 505 \(\Omega\)
(D) 5 \(\Omega\)
Correct Answer: (B) 502.5 \(\Omega\)
View SolutionStep 1: Understanding the Concept:
A Wheatstone bridge is balanced when the ratio of the two resistors in one arm equals the ratio in the other arm.
Step 2: First balance:
With P and Q in place, balance gives \[ \frac{P}{Q} = \frac{R_1}{S} = \frac{500}{S} \]
Step 3: Second balance:
After P and Q are interchanged, the ratio flips. So \[ \frac{Q}{P} = \frac{R_2}{S} = \frac{505}{S} \]
Step 4: Multiply the two results:
Multiplying gives \[ 1 = \frac{500 \times 505}{S^2} \] \[ S = \sqrt{500 \times 505} = \sqrt{252500} \approx 502.49\ \Omega \]
Step 5: Compare with the options:
The value is about 502.5 ohm. 500 and 505 are the two balance values, and S is between them. 5 ohm is the difference and has no meaning here.
Final Answer:
S is the geometric mean of 500 and 505, about 502.5 ohm.
\[ \boxed{S \approx 502.5\ \Omega} \]
Question 13:
A positive charge enters a magnetic field and travels parallel to the magnetic field. The charge experiences
(A) an upward force
(B) a downward force
(C) a force in the direction of magnetic field.
(D) no force
Correct Answer: (D) no force
View SolutionStep 1: Understanding the Concept:
A moving charge in a magnetic field feels the Lorentz force \(\vec{F} = q(\vec{v}\times\vec{B})\).
Step 2: Apply to parallel motion:
The velocity is parallel to the field, so the angle between them is \(\theta = 0^\circ\). The magnitude is \[ F = qvB\sin 0^\circ = 0 \]
Step 3: Why the other options fail:
Any force from a magnetic field is perpendicular to both \(\vec{v}\) and \(\vec{B}\). So a force along the field (option 3) is never possible. An upward or downward force (options 1 and 2) needs a nonzero cross product, which is zero here.
Final Answer:
The charge feels no force.
\[ \boxed{\text{No force}} \]
Question 14:
If a ferromagnetic material is inserted in a current carrying solenoid, the magnetic field inside the solenoid
(A) increases
(B) decreases
(C) remains unchanged
(D) becomes zero
Correct Answer: (A) increases
View SolutionStep 1: Understanding the Concept:
The field inside an air-filled solenoid is \(B_0 = \mu_0 n I\). With a core of relative permeability \(\mu_r\), the field becomes \(B = \mu_r \mu_0 n I\).
Step 2: Ferromagnetic core:
For ferromagnetic materials such as iron, \(\mu_r\) is very large, from hundreds to thousands. So \(B\) becomes much larger than \(B_0\).
Step 3: Why the other options fail:
The field cannot fall or vanish because \(\mu_r > 1\). It cannot stay unchanged because the domains of the core align with the field and add their own magnetisation.
Final Answer:
The magnetic field inside the solenoid increases.
\[ \boxed{\text{Increases}} \]
Question 15:
An electron of charge 'e' moves in a circular orbit of radius r around a nucleus with a frequency \(\nu\). The magnetic moment associated with the orbital motion of the electron is
(A) \(\pi \nu e r^2\)
(B) \(\dfrac{\pi \nu r^2}{e}\)
(C) \(\dfrac{\pi \nu e}{r}\)
(D) \(\dfrac{\pi e r^2}{\nu}\)
Correct Answer: (A) \(\pi \nu e r^2\)
View SolutionStep 1: Understanding the Concept:
A charge going round a loop forms a tiny current loop. Its magnetic moment is \(M = I A\).
Step 2: Find the current:
The electron passes a point \(\nu\) times each second, so the current magnitude is \(I = e\nu\).
Step 3: Find the area:
The orbit is a circle, so \(A = \pi r^2\).
Step 4: Multiply:
\[ M = I A = e\nu \times \pi r^2 = \pi \nu e r^2 \]
Step 5: Check the options:
Options 2 and 4 divide by e or \(\nu\), which does not match \(I = e\nu\). Option 3 has r in the denominator and no \(r^2\). Only option 1 fits.
Final Answer:
The orbital magnetic moment is \(\pi \nu e r^2\).
\[ \boxed{M = \pi \nu e r^{2}} \]
Question 16:
Identify the correct statements among the following. A. The torque acting on a planar current loop in a magnetic field changes when its shape is changed without changing its area. B. Two parallel wires carrying currents in opposite directions attract each other. C. An ideal ammeter has zero resistance and an ideal voltmeter has infinite resistance. D. A galvanometer can be converted into a voltmeter by connecting a high resistance in series.
Choose the correct answer from the options given below:
(A) A, B and C only
(B) B, C and D only
(C) C and D only
(D) A and C only
Correct Answer: (C) C and D only
View SolutionStep 1: Understand the question.
We have to judge four statements about current loops, parallel wires and meters. Then we pick the option that lists only the true ones.
Step 2: Check statement (A).
The torque on a planar loop is \(\tau = NIAB\sin\theta\). It depends only on the current, the area and the field, not on the shape. If the area stays the same, the torque stays the same. So (A) is FALSE.
Step 3: Check statement (B).
Parallel wires with currents in the same direction attract. Wires with currents in opposite directions repel. So (B) is FALSE.
Step 4: Check statement (C).
An ammeter is joined in series, so it should add no extra resistance. An ideal ammeter has zero resistance. A voltmeter is joined in parallel, so it should draw no current. An ideal voltmeter has infinite resistance. So (C) is TRUE.
Step 5: Check statement (D).
A galvanometer becomes a voltmeter when a large resistance \(R = \frac{V}{I_g} - G\) is put in series with it. This keeps the current through the coil small. So (D) is TRUE.
Step 6: Match with the options.
Only C and D are true, which is option 3. Options 1 and 2 include B, and options 1 and 4 include A, so they are wrong.
Final Answer:
Statements C and D are correct, so option 3 is right.
\[ \boxed{\text{Option 3: C and D only}} \]
Question 17:
A 0.5 m long solenoid has 500 turns and has a flux density of \(2.52 \times 10^{-3}\) T at its center. The current in the solenoid is (Given, \(\mu_0 = 4\pi \times 10^{-7}\) H/m)
(A) 1.2 A
(B) 2.0 A
(C) 2.8 A
(D) 3.4 A
Correct Answer: (B) 2.0 A
View SolutionStep 1: Understanding the concept.
The field inside a long solenoid is uniform. It depends on the number of turns per unit length and on the current.
Step 2: Key formula.
\[ B = \mu_0 n I \] where \(n = N/L\) is the number of turns per metre.
Step 3: Find n.
\[ n = \frac{500}{0.5} = 1000 \text{ turns/m} \]
Step 4: Solve for I.
\[ I = \frac{B}{\mu_0 n} = \frac{2.52 \times 10^{-3}}{4\pi \times 10^{-7} \times 1000} = \frac{2.52 \times 10^{-3}}{1.2566 \times 10^{-3}} \approx 2.0 \text{ A} \]
Step 5: Check the options.
1.2 A, 2.8 A and 3.4 A do not match the value 2.005 A. Only 2.0 A matches, which is option 2.
Final Answer:
The current in the solenoid is 2.0 A.
\[ \boxed{2.0 \text{ A}} \]
Question 18:
A bar magnet of dipole moment 3 Am2 rests with its centre on a frictionless point. A force F is applied at right angles to the axis of the magnet, 10 cm from the point. It is observed that an external magnetic field of 0.25 T is required to hold the magnet in equilibrium at an angle of \(30^{\circ}\) with the field. The value of F is:
(A) 2.75 N
(B) 2.93 N
(C) 3.75 N
(D) 4.08 N
Correct Answer: (C) 3.75 N
View SolutionStep 1: Understanding the concept.
The field tries to turn the magnet with a torque. The applied force F gives an opposite torque about the pivot. For equilibrium the two torques are equal.
Step 2: Torque by the field.
\[ \tau_B = MB\sin\theta = 3 \times 0.25 \times \sin 30^{\circ} = 3 \times 0.25 \times 0.5 = 0.375 \text{ N m} \]
Step 3: Torque by the force.
F acts at right angles to the axis at distance 0.10 m from the pivot, so \[ \tau_F = F \times 0.10 \]
Step 4: Balance the torques.
\[ F \times 0.10 = 0.375 \Rightarrow F = 3.75 \text{ N} \]
Step 5: Check the options.
2.75 N, 2.93 N and 4.08 N do not satisfy the balance. 3.75 N is option 3.
Final Answer:
F = 3.75 N.
\[ \boxed{3.75 \text{ N}} \]
Question 19:
A wire shown in figure carries a current of 10 A. The magnitude of the magnetic field at the centre O is: (Given: radius of the bent coil is 3 cm.)
(A) \(1 \times 10^{-4}\) T
(B) \(1.57 \times 10^{-3}\) T
(C) \(1.57 \times 10^{-4}\) T
(D) \(2.41 \times 10^{-5}\) T
Correct Answer: (C) \(1.57 \times 10^{-4}\) T
View SolutionStep 1: Understanding the figure.
The wire has two straight parts and a circular arc. The dashed lines show that the straight wires, if extended, pass through the centre O. The arc covers \(360^{\circ} - 90^{\circ} = 270^{\circ}\), which is \(\frac{3}{4}\) of a circle.
Step 2: Straight parts.
A point that lies on the line of a straight current has \(d\vec{l} \times \vec{r} = 0\). So the two straight wires give zero field at O.
Step 3: Arc part.
A full circle gives \(\frac{\mu_0 I}{2r}\) at the centre. Three quarters of it gives \[ B = \frac{3}{4}\cdot\frac{\mu_0 I}{2r} = \frac{3}{4} \times \frac{4\pi \times 10^{-7} \times 10}{2 \times 0.03} \]
Step 5: Check the options.
\(1.57 \times 10^{-3}\) T is ten times too large. \(1 \times 10^{-4}\) T and \(2.41 \times 10^{-5}\) T do not match. The right value is option 3.
Final Answer:
The field at O is \(1.57 \times 10^{-4}\) T.
\[ \boxed{1.57 \times 10^{-4} \text{ T}} \]
Question 20:
The mutual inductance of a pair of coils placed close to each other depends upon
(A) The rate at which currents are changing in the two coils.
(B) Relative position and orientation of the two coils
(C) The materials of the wires of the coils
(D) The current in the two coils
Correct Answer: (B) Relative position and orientation of the two coils
View SolutionStep 1: Understanding the concept.
Mutual inductance M links the emf in one coil to the changing current in the other, through \(\varepsilon = -M\frac{dI}{dt}\). It is a geometric property of the pair.
Step 2: Check option (1).
The rate of change of current sets the emf, not M itself. M is the same whatever the rate. So option 1 is wrong.
Step 3: Check option (2).
M depends on how much flux of one coil links the other. That depends on the distance, the number of turns, the area, the medium and the relative orientation of the coils. So option 2 is right.
Step 4: Check option (3).
The wire material only changes resistance, not the flux linkage. So option 3 is wrong.
Step 5: Check option (4).
M is defined per unit current, so it does not depend on the current value. Option 4 is wrong.
Final Answer:
Mutual inductance depends on the relative position and orientation of the coils.
\[ \boxed{\text{Option 2}} \]
Question 21:
If the number of turns in a coil is tripled, the value of the magnetic flux linked with the coil:
(A) remains unchanged
(B) becomes \(\frac{1}{3}\) times
(C) is tripled
(D) becomes \(\frac{2}{3}\) times
Correct Answer: (C) is tripled
View SolutionStep 1: Understanding the concept.
Total flux linked with a coil of N turns is \(\phi = NBA\cos\theta\). Every turn links the same flux, and the turns add up.
Step 2: Apply to the question.
B, A and the angle stay the same. Only N changes from N to 3N.
Step 4: Check the options.
Option 1 ignores N. Options 2 and 4 make the flux smaller, which cannot happen when turns increase. Option 3 is right.
Final Answer:
The flux linked is tripled.
\[ \boxed{\text{Option 3: is tripled}} \]
Question 22:
Identify the correct statement(s) among the following. A. In a d.c. circuit, a capacitor can conduct but not an inductor. B. In a d.c. circuit, an inductor can conduct but not a capacitor. C. In a d.c. circuit, both the inductor and capacitor cannot conduct. D. An inductor has infinite resistance in a d.c. circuit.
Choose the correct answer from the options given below:
(A) A and C only
(B) B and D only
(C) B only
(D) A, C and D only
Correct Answer: (C) B only
View SolutionStep 1: Understanding the concept.
For steady d.c. the frequency is zero. Inductive reactance is \(X_L = \omega L = 0\). Capacitive reactance is \(X_C = \frac{1}{\omega C} \to \infty\).
Step 2: Check statement (A).
A capacitor blocks steady d.c. after it charges, and an inductor passes it. So A says the reverse of the truth. FALSE.
Step 3: Check statement (B).
The inductor passes d.c. freely (it acts like a plain wire) and the capacitor blocks it. So (B) is TRUE.
Step 4: Check statement (C).
The inductor does conduct d.c., so it is wrong to say both cannot. (C) is FALSE.
Step 5: Check statement (D).
The inductor has zero reactance in d.c., not infinite resistance. (D) is FALSE.
Step 6: Match with options.
Only B is true, which is option 3.
Final Answer:
Only statement B is correct.
\[ \boxed{\text{Option 3: B only}} \]
Question 23:
An a.c. voltage is applied to a pure inductor. The current in the inductor would be
(A) leading the voltage by \(\frac{\pi}{2}\)
(B) lagging the voltage by \(\frac{\pi}{2}\)
(C) leading the voltage by \(\frac{\pi}{4}\)
(D) in phase with the voltage.
Correct Answer: (B) lagging the voltage by \(\frac{\pi}{2}\)
View SolutionStep 1: Understanding the Concept:
In a pure inductor there is no resistance. The only opposition to the current comes from the self-induced emf, which opposes any change in current.
Step 2: Key Formula or Approach:
Let the applied voltage be \(V = V_0 \sin \omega t\). The induced emf balances the applied voltage, so \(V = L \frac{dI}{dt}\).
Step 3: Detailed Explanation:
Integrate the relation to get the current.
\[ I = \frac{1}{L}\int V_0 \sin \omega t \, dt = -\frac{V_0}{\omega L}\cos \omega t \]
\[ I = \frac{V_0}{\omega L}\sin\left(\omega t - \frac{\pi}{2}\right) \]
The current has a phase of \(-\pi/2\) compared with the voltage. So the current reaches its peak a quarter cycle after the voltage does.
Step 4: Check the options:
Option 1 says the current leads. That is true for a capacitor, not an inductor. Option 3 gives a lead of \(\pi/4\), which happens for no pure element. Option 4 is true only for a pure resistor. Option 2 matches our result.
Final Answer:
In a pure inductor the current lags the voltage by \(\pi/2\).
\[ \boxed{\text{Option 2: lagging the voltage by } \frac{\pi}{2}} \]
Question 24:
A pure inductor of 0.25 H is connected to a source of 220 V. If the frequency of the source is 50 Hz, then the rms current in the circuit will be.
(A) 1.6 A
(B) 4.0 A
(C) 2.8 A
(D) 28 A
Correct Answer: (C) 2.8 A
View SolutionStep 1: Understanding the Concept:
A pure inductor opposes an a.c. current through its reactance. The rms current equals the rms voltage divided by this reactance.
Step 2: Key Formula or Approach:
Inductive reactance is \(X_L = 2\pi f L\). Then \(I_{rms} = \frac{V_{rms}}{X_L}\).
Step 5: Check the options:
1.6 A would need a reactance of about 137 ohm. 4.0 A would need 55 ohm. 28 A would need 7.85 ohm, which is off by a factor of 10 and comes from a decimal slip. Only 2.8 A fits.
Final Answer:
The rms current is about 2.8 A, which is option 3.
\[ \boxed{I_{rms} \approx 2.8\ \text{A}} \]
Question 25:
A coil with an average diameter of 0.02 m is placed with its plane perpendicular to a magnetic field of 6000 T. The induced emf in the coil is 11 V, when the magnetic field is changed to 1000 T in 4 s. The number of turns in the coil is
(A) 16
(B) 18
(C) 24
(D) 28
Correct Answer: (D) 28
View SolutionStep 1: Understanding the Concept:
By Faraday's law the emf induced in a coil of N turns equals N times the rate of change of flux through one turn.
Step 2: Key Formula or Approach:
\(\varepsilon = N \frac{A\,|\Delta B|}{\Delta t}\), because the plane is perpendicular to the field and the flux per turn is \(BA\).
Step 3: Find the area:
Radius is \(r = 0.01\ \text{m}\). So \(A = \pi r^2 = \pi \times 10^{-4} \approx 3.14 \times 10^{-4}\ \text{m}^2\).
Step 4: Find the change in field:
\(|\Delta B| = 6000 - 1000 = 5000\ \text{T}\), and \(\Delta t = 4\ \text{s}\). So \(\frac{\Delta B}{\Delta t} = 1250\ \text{T/s}\).
Step 6: Check the options:
With N = 16, 18 or 24 the emf would be 6.3 V, 7.1 V or 9.4 V, so none matches 11 V. N = 28 gives \(28 \times 0.3927 = 11.0\) V.
Final Answer:
The coil has 28 turns.
\[ \boxed{N = 28} \]
Question 26:
Arrange the following electromagnetic waves in the ascending order of their energies. A. X-rays B. UV-rays C. Infrared rays D. Microwaves Choose the correct answer from the options given below:
(A) A, B, C, D
(B) B, C, D, A
(C) D, B, A, C
(D) D, C, B, A
Correct Answer: (D) D, C, B, A
View SolutionStep 1: Understanding the Concept:
The energy of a photon is \(E = h\nu = hc/\lambda\). So higher frequency means higher energy, and longer wavelength means lower energy.
Step 2: Recall the spectrum order:
In order of increasing frequency the spectrum runs: radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays.
Step 3: Pick the four given waves:
Lowest energy is microwaves (D). Next is infrared (C). Then ultraviolet (B). The highest is X-rays (A).
Step 4: Write the order:
Ascending order of energy is D, C, B, A.
Step 5: Check the options:
Option 1 is the descending order. Option 2 puts X-rays last but misplaces microwaves. Option 3 wrongly puts infrared after X-rays. Option 4 is correct.
Final Answer:
The ascending order is microwaves, infrared, ultraviolet, X-rays.
\[ \boxed{D, C, B, A} \]
Question 27:
The conduction current is same as displacement current in a circuit when the source is
(A) a.c. only
(B) d.c. only
(C) either a.c. or d.c.
(D) neither d.c nor a.c.
Correct Answer: (C) either a.c. or d.c.
View SolutionStep 1: Understanding the Concept:
Maxwell added the displacement current \(I_d = \varepsilon_0 \frac{d\Phi_E}{dt}\) so that the total current stays continuous across a capacitor gap.
Step 2: Key Idea:
Between the plates there is no conduction current, but the electric field changes. Whenever charge flows in the wires, the field between the plates changes at the matching rate, so \(I_d = I_c\).
Step 3: Apply to an a.c. source:
With an a.c. source the capacitor charges and discharges all the time. The conduction current in the wires equals the displacement current in the gap at every instant.
Step 4: Apply to a d.c. source:
With a d.c. source the capacitor charges while the current flows in the wires. During this time the conduction current again equals the displacement current. After full charge both are zero, so they are still equal.
Step 5: Check the options:
The equality does not depend on the type of source, so it holds for a.c. and for d.c. Options 1 and 2 wrongly restrict it to one type. Option 4 is false because they are equal in both cases.
Final Answer:
The two currents match for either kind of source.
\[ \boxed{\text{Option 3: either a.c. or d.c.}} \]
Question 28:
Identify the correct statements among the following. A. All electromagnetic waves, irrespective of their frequencies, travel at the same speed through vacuum. B. A capacitor is fully charged by a d.c. source. When it is fully charged, the displacement current is zero. C. Ultraviolet radiation can be used for viewing objects through haze and fog. D. A moving charge in a circular orbit cannot produce electromagnetic waves. Choose the correct answer from the options given below:
(A) A, B and C only
(B) B and C only
(C) A and B only
(D) C and D only
Correct Answer: (C) A and B only
View SolutionStep 1: Understanding the Concept:
Each statement tests a basic idea about electromagnetic waves. Check them one by one.
Step 2: Check statement (A):
In vacuum every electromagnetic wave travels at \(c = 1/\sqrt{\mu_0\varepsilon_0} \approx 3\times10^8\) m/s, whatever its frequency. So (A) is TRUE.
Step 3: Check statement (B):
Once a capacitor is fully charged by a d.c. source, the electric field between the plates stops changing. Then \(d\Phi_E/dt = 0\), so the displacement current is zero. So (B) is TRUE.
Step 4: Check statement (C):
Haze and fog particles scatter short wavelengths strongly. UV scatters far more than infrared, so it cannot pass through. Infrared is used to see through haze and fog. So (C) is FALSE.
Step 5: Check statement (D):
A charge moving in a circle is accelerating, because its velocity direction keeps changing. An accelerated charge radiates electromagnetic waves. So (D) is FALSE.
Step 6: Pick the option:
Only A and B are correct, which is option 3.
Final Answer:
Statements A and B are correct.
\[ \boxed{A \text{ and } B \text{ only}} \]
Question 29:
A radiation of energy \(E\) falls normally on a perfectly reflecting surface. The momentum transferred to the surface is
(A) \(\frac{2E}{c}\)
(B) \(\frac{2E}{c^2}\)
(C) \(\frac{E}{c^2}\)
(D) \(\frac{E}{c}\)
Correct Answer: (A) \(\frac{2E}{c}\)
View SolutionStep 1: Understanding the Concept:
Electromagnetic radiation carries momentum as well as energy. For radiation of energy \(E\), the momentum is \(p = E/c\).
Step 2: Key Formula or Approach:
Momentum transferred equals the change in momentum of the radiation. For total reflection at normal incidence the direction reverses.
Step 3: Detailed Explanation:
Initial momentum of radiation: \(+E/c\). Final momentum after reflection: \(-E/c\). The change is
\[ \Delta p = \frac{E}{c} - \left(-\frac{E}{c}\right) = \frac{2E}{c} \]
By conservation of momentum the surface gains \(2E/c\).
Step 4: Check the options:
\(E/c\) is the result for a perfect absorber, where the radiation only stops. \(E/c^2\) and \(2E/c^2\) have the wrong units for momentum. Only \(2E/c\) fits.
Final Answer:
The surface gets a momentum of \(2E/c\).
\[ \boxed{\frac{2E}{c}} \]
Question 30:
The intensity of a laser beam is \(2.5 \times 10^{14}\ \text{W/m}^2\). The amplitude of the magnetic field in the beam is (Given, \(\epsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}\))
(A) ~1.15 T
(B) ~1.45 T
(C) ~2.0 T
(D) ~3.25 T
Correct Answer: (B) ~1.45 T
View SolutionStep 1: Understanding the Question:
A laser beam is an electromagnetic wave. We know its intensity and need the amplitude of the magnetic field \(B_0\). The speed of light is \(c = 3 \times 10^{8}\ \text{m/s}\).
Step 2: Key Formula or Approach:
The average intensity of an electromagnetic wave is the energy carried per second per unit area. It is given by \[ I = \frac{1}{2}\,\epsilon_0 E_0^2\, c \] The two field amplitudes are linked by \(E_0 = cB_0\).
Step 3: Find the Electric Field Amplitude:
Rearrange the intensity formula. \[ E_0 = \sqrt{\frac{2I}{\epsilon_0 c}} = \sqrt{\frac{2 \times 2.5 \times 10^{14}}{8.85 \times 10^{-12} \times 3 \times 10^{8}}} \] The denominator is \(2.655 \times 10^{-3}\). So \[ E_0 = \sqrt{1.883 \times 10^{17}} \approx 4.34 \times 10^{8}\ \text{V/m} \]
Step 4: Find the Magnetic Field Amplitude:
Use \(B_0 = E_0/c\). \[ B_0 = \frac{4.34 \times 10^{8}}{3 \times 10^{8}} \approx 1.45\ \text{T} \]
Step 5: Checking Each Option:
Option 1 (1.15 T) and option 3 (2.0 T) do not match the calculated 1.45 T. Option 4 (3.25 T) is far too large. Only option 2 matches.
Final Answer:
The amplitude of the magnetic field is about 1.45 T, so option 2 is correct.
\[ \boxed{B_0 \approx 1.45\ \text{T}} \]
Question 31:
A real image is formed by a convex lens on the screen. If half of the lens is covered by an opaque object, then
(A) half image is not seen
(B) full image of same intensity is seen
(C) half image of same intensity is seen
(D) full image of decreased intensity is seen
Correct Answer: (D) full image of decreased intensity is seen
View SolutionStep 1: Understanding the Question:
A convex lens forms a real image on a screen. Half of the lens is blocked. We must say what happens to the image.
Step 2: Key Idea:
Light from every point of the object spreads out and falls on the whole face of the lens. Each part of the lens sends light from that object point to the same image point. So every part of the lens forms the complete image.
Step 3: What Blocking Half the Lens Does:
If half the lens is covered, the other half still receives light from every point of the object. It still forms the full image. But only half the light gets through, so the image is dimmer.
Step 4: Checking Each Option:
Option 1 says half the image is lost. This is wrong because the remaining half of the lens still forms the whole image. Option 2 says the intensity stays the same. This is wrong because only half the light passes. Option 3 is wrong for both reasons. Option 4 says the full image is seen with less intensity. This is correct.
Final Answer:
The image stays complete but becomes dimmer, so option 4 is correct.
\[ \boxed{\text{Full image of decreased intensity}} \]
Question 32:
Arrange the following substances/media in the increasing order of their refractive indices. A. Crown glass B. Water C. Diamond D. CO2 Choose the correct answer from the options given below:
(A) A, B, C, D
(B) D, C, B, A
(C) D, B, A, C
(D) D, B, C, A
Correct Answer: (C) D, B, A, C
View SolutionStep 1: Understanding the Question:
We must sort four media by refractive index, from the smallest to the largest.
Step 2: Recall the Values:
Refractive index is \(n = c/v\). Typical values are: CO\(_2\) gas about 1.0004, water about 1.33, crown glass about 1.52, diamond about 2.42.
Step 3: Sort in Increasing Order:
\[ n_{CO_2} < n_{water} < n_{crown\ glass} < n_{diamond} \] In letters this is D, B, A, C.
Step 4: Checking Each Option:
Option 1 (A, B, C, D) puts crown glass first. This is wrong. Option 2 (D, C, B, A) puts diamond before water. This is wrong. Option 3 (D, B, A, C) matches our order. Option 4 (D, B, C, A) puts diamond before crown glass. This is wrong.
Final Answer:
The order is CO2, water, crown glass, diamond, so option 3 is correct.
\[ \boxed{D, B, A, C} \]
Question 33:
Match the LIST-I with LIST-II (In the context of Young's double slit experiment)
LIST-I
LIST-II
A. The width of one slit is slightly increased.
I. The fringe width increases.
B. One slit is closed.
II. Interference pattern becomes less sharp.
C. The width of the source slit is increased.
III. Maximum intensity increases
D. Light of smaller frequency is used.
IV. Interference pattern disappears
Choose the correct answer from the options given below:
(A) A-I, B-II, C-III, D-IV
(B) A-IV, B-III, C-II, D-I
(C) A-III, B-IV, C-II, D-I
(D) A-III, B-IV, C-I, D-II
Correct Answer: (C) A-III, B-IV, C-II, D-I
View SolutionStep 1: Understanding the Question:
We match each change in Young's double slit setup with its effect on the pattern.
Step 2: Match A:
When one slit is made a little wider, it lets more light through. The total light reaching the screen rises, so the maximum intensity increases. So A matches III.
Step 3: Match B:
If one slit is closed, there is only one source of light on the screen. There are no two waves to interfere. So the interference pattern disappears. B matches IV.
Step 4: Match C:
A wider source slit acts like many nearby sources. Each one makes its own fringe system, and these are slightly shifted. They overlap and wash out the dark and bright bands. So the pattern becomes less sharp. C matches II.
Step 5: Match D:
Fringe width is \(\beta = \lambda D / d\). Smaller frequency means larger wavelength, since \(\lambda = c/f\). So \(\beta\) increases. D matches I.
Step 6: Choose the Option:
We have A-III, B-IV, C-II, D-I. This is the third printed option. The other options break at least one of these matches.
Final Answer:
The matching is A-III, B-IV, C-II, D-I, so option 3 is correct.
\[ \boxed{\text{A-III, B-IV, C-II, D-I}} \]
Question 34:
In a Young's double slit experiment, the intensity at the center of the screen is I. If one of the slits is closed, the intensity at the center of the screen now will be:
(A) \(\dfrac{1}{2}\)
(B) 1
(C) \(\dfrac{1}{4}\)
(D) \(\dfrac{1}{3}\)
Correct Answer: (C) \(\dfrac{1}{4}\)
View SolutionStep 1: Understanding the Question:
At the center of the screen the path difference is zero. Both waves arrive in phase. When one slit is closed, only one wave remains.
Step 2: Key Formula or Approach:
If each slit alone gives intensity \(I_0\), the two waves in phase have amplitude \(2a\), where \(a\) is the amplitude of one wave. Intensity goes as the square of the amplitude.
Step 3: Both Slits Open:
\[ I = k(2a)^2 = 4ka^2 = 4I_0 \] So \(I_0 = I/4\).
Step 4: One Slit Closed:
Only one wave of amplitude \(a\) reaches the center. The intensity is \[ I' = ka^2 = I_0 = \frac{I}{4} \]
Step 5: Checking Each Option:
Option 1 (I/2) is what you would get if intensity were linear in amplitude. That is wrong. Option 2 says no change. Wrong. Option 4 (I/3) has no basis. Option 3 (I/4) matches.
Final Answer:
The intensity at the center becomes one fourth of I, so option 3 is correct.
\[ \boxed{\frac{I}{4}} \]
Question 35:
A ray of light of frequency \(5 \times 10^{14}\) Hz is passed through a liquid. The wavelength of light measured inside the liquid is found to be \(450 \times 10^{-9}\) m. The refractive index of the liquid is
(A) 1.25
(B) 1.33
(C) 1.45
(D) 1.51
Correct Answer: (B) 1.33
View SolutionStep 1: Understanding the Question:
Light of a fixed frequency enters a liquid. We know the wavelength inside the liquid. We must find the refractive index.
Step 2: Key Formula or Approach:
Frequency does not change when light enters a new medium. Speed in the medium is \(v = f\lambda\). The refractive index is \(n = c/v\).
Step 3: Find the Speed in the Liquid:
\[ v = f\lambda = 5 \times 10^{14} \times 450 \times 10^{-9} = 2.25 \times 10^{8}\ \text{m/s} \]
Step 5: Checking Each Option:
1.25, 1.45 and 1.51 do not match 1.33. Option 2 is correct.
Final Answer:
The refractive index of the liquid is 1.33, so option 2 is correct.
\[ \boxed{n = 1.33} \]
Question 36:
If the two slits in a Young's double slit experiment have their widths in the ratio 4:1, then the ratio of intensities at maxima and minima in the interference pattern will be
(A) 16 : 3
(B) 16 : 1
(C) 25 : 9
(D) 9 : 1
Correct Answer: (D) 9 : 1
View SolutionStep 1: Understanding the Question:
The slit widths are in the ratio 4:1. We must find the ratio \(I_{max}/I_{min}\).
Step 2: Key Formula or Approach:
The intensity from a slit is proportional to its width. So \(I_1 : I_2 = 4 : 1\). Amplitude goes as the square root of intensity, so \(a_1 : a_2 = 2 : 1\). \[ I_{max} \propto (a_1 + a_2)^2, \qquad I_{min} \propto (a_1 - a_2)^2 \]
Step 4: Checking Each Option:
25:9 comes from wrongly using the width ratio 4:1 as the amplitude ratio. 16:1 comes from squaring the width ratio. 16:3 has no basis. Option 4 (9:1) is correct.
Final Answer:
The ratio of maximum to minimum intensity is 9 : 1, so option 4 is correct.
\[ \boxed{9 : 1} \]
Question 37:
A concave mirror forms a real image of an object kept at a distance of 9 cm from it. If the object is taken away further from the mirror by 6 cm, the image size is reduced to \((1/4)^{th}\) of its previous size. The focal length of the mirror is:
(A) -5 cm
(B) -7 cm
(C) 15 cm
(D) 18 cm
Correct Answer: (B) -7 cm
View SolutionStep 1: Understanding the Question:
A concave mirror makes a real image of an object at 9 cm. The object is then moved to 15 cm, and the image size becomes one fourth. We must find the focal length.
Step 2: Key Formula or Approach:
The magnification is \(m = \dfrac{f}{f-u}\), with distances measured from the mirror and the object distance \(u\) negative. For a concave mirror \(f\) is negative.
Step 3: Write Both Magnifications:
First position: \(u_1 = -9\). \[ m_1 = \frac{f}{f+9} \] Second position: \(u_2 = -15\). \[ m_2 = \frac{f}{f+15} \] The image size is one fourth, so \(m_2 = \tfrac{1}{4}m_1\).
Step 4: Solve for f:
Both images are real, so both magnifications are negative and their ratio is positive. \[ \frac{f+9}{f+15} = \frac{1}{4} \] \[ 4f + 36 = f + 15 \quad\Rightarrow\quad 3f = -21 \quad\Rightarrow\quad f = -7\ \text{cm} \]
Step 5: Check:
With \(f = -7\): \(m_1 = -7/2 = -3.5\) and \(m_2 = -7/8 = -0.875\). The ratio is \(0.875/3.5 = 1/4\). The object at 9 cm is beyond the 7 cm focal length, so the image is real.
Step 6: Checking Each Option:
-5 cm gives \(m_1 = -5/4\) and \(m_2 = -5/10\), so \(m_2/m_1 = 0.4\), not one fourth. Positive focal lengths (15 cm, 18 cm) mean a convex mirror, which cannot form a real image. So option 2 is correct.
Final Answer:
The focal length of the mirror is -7 cm, so option 2 is correct.
\[ \boxed{f = -7\ \text{cm}} \]
Question 38:
Arrange the following photosensitive metals (work function given in brackets) according to their threshold wavelengths in decreasing order of their values. A. K (\(\Phi\) = 2.3 eV) B. Na (\(\Phi\) = 2.75 eV) C. Cs (\(\Phi\) = 2.14 eV) D. Ca (\(\Phi\) = 3.2 eV)
Choose the correct answer from the options given below:
(A) A, B, C, D
(B) D, C, B, A
(C) C, A, D, B
(D) C, A, B, D
Correct Answer: (D) C, A, B, D
View SolutionStep 1: Understanding the Question:
The threshold wavelength is the longest wavelength of light that can still knock electrons out of a metal. We must list the four metals from the largest threshold wavelength to the smallest.
Step 2: Key Formula or Approach:
The work function is the minimum energy needed to free an electron. It is linked to the threshold wavelength by \[ \Phi = \frac{hc}{\lambda_0} \quad \Rightarrow \quad \lambda_0 = \frac{hc}{\Phi} \] Here h and c are constants, so \(\lambda_0\) is inversely proportional to \(\Phi\).
Step 3: Decide the Order:
A larger work function gives a smaller threshold wavelength. So the largest threshold wavelength belongs to the smallest work function. Work functions in increasing order: Cs (2.14 eV) < K (2.3 eV) < Na (2.75 eV) < Ca (3.2 eV). So threshold wavelengths in decreasing order: Cs > K > Na > Ca, which is C, A, B, D.
Step 4: Checking Each Option:
Option 1 (A, B, C, D) is K, Na, Cs, Ca. Cs has the smallest work function, so it cannot come third. Wrong. Option 2 (D, C, B, A) puts Ca first, but Ca has the largest work function and the shortest threshold wavelength. Wrong. Option 3 (C, A, D, B) puts Ca before Na, but Na has the smaller work function. Wrong. Option 4 (C, A, B, D) matches our order. Correct.
Final Answer:
The order of decreasing threshold wavelength is Cs, K, Na, Ca, that is C, A, B, D. This is option 4.
\[ \boxed{4} \]
Question 39:
An electron and a photon each have the same de-Broglie wavelength of 1.0 nm. The ratio of their linear momenta is
(A) 1:1
(B) 1:2
(C) 2:1
(D) 4:9
Correct Answer: (A) 1:1
View SolutionStep 1: Understanding the Question:
Both particles have the same wavelength, 1.0 nm. We must compare their momenta as a ratio.
Step 2: Key Formula or Approach:
The de-Broglie relation for any particle, and also the relation for a photon, is \[ \lambda = \frac{h}{p} \quad \Rightarrow \quad p = \frac{h}{\lambda} \]
Step 3: Momentum of the Electron:
\[ p_e = \frac{h}{\lambda} \]
Step 4: Momentum of the Photon:
For a photon, \(E = hc/\lambda\) and \(p = E/c = h/\lambda\). So \[ p_{ph} = \frac{h}{\lambda} \]
Step 5: Take the Ratio:
\[ \frac{p_e}{p_{ph}} = \frac{h/\lambda}{h/\lambda} = 1 \] The ratio is 1:1. The mass and speed of the two particles are very different, but momentum depends only on wavelength.
Step 6: Checking Each Option:
Option 1 (1:1) is correct. Options 2, 3 and 4 would need the momenta to differ, which the relation \(p = h/\lambda\) does not allow when \(\lambda\) is equal.
Final Answer:
Same wavelength means same momentum, so the ratio is 1:1. This is option 1.
\[ \boxed{1:1} \]
Question 40:
A monochromatic source emitting light of 600 nm, has a power output of 66 W. The number of photons emitted by the source per second is (Given, Plank's constant h = 6.6 \(\times\) 10\(^{-34}\) Js)
(A) 2 \(\times\) 10\(^{18}\)
(B) 2 \(\times\) 10\(^{19}\)
(C) 2 \(\times\) 10\(^{20}\)
(D) 8 \(\times\) 10\(^{22}\)
Correct Answer: (C) 2 \(\times\) 10\(^{20}\)
View SolutionStep 1: Understanding the Question:
A source gives out 66 J of light energy every second. Each photon carries a fixed energy that depends on the wavelength. We need the number of photons per second.
Step 2: Key Formula or Approach:
Energy of one photon: \(E = \dfrac{hc}{\lambda}\). Power is total energy per second, so \[ P = n E \quad \Rightarrow \quad n = \frac{P}{E} = \frac{P\lambda}{hc} \] where n is the number of photons per second.
Step 3: Energy of One Photon:
\(\lambda = 600 \text{ nm} = 6 \times 10^{-7}\) m and \(c = 3 \times 10^{8}\) m/s. \[ E = \frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{6 \times 10^{-7}} = 3.3 \times 10^{-19} \text{ J} \]
Step 4: Number of Photons:
\[ n = \frac{66}{3.3 \times 10^{-19}} = 2 \times 10^{20} \text{ per second} \]
Step 5: Checking Each Option:
Option 1 (\(2 \times 10^{18}\)) and option 2 (\(2 \times 10^{19}\)) are too small by factors of 100 and 10. Option 4 (\(8 \times 10^{22}\)) is far too large. Option 3 (\(2 \times 10^{20}\)) matches our result.
Final Answer:
The source emits \(2 \times 10^{20}\) photons every second, which is option 3.
\[ \boxed{2 \times 10^{20}} \]
Question 41:
Match the LIST-I with LIST-II (Symbols have their usual meaning)
LIST-I Spectral series of Hydrogen
LIST-II Wave Number
A. Lyman series
I. \(\bar{v} = R\left[\dfrac{1}{1^2} - \dfrac{1}{n_2^2}\right]\)
B. Paschen series
II. \(\bar{v} = R\left[\dfrac{1}{2^2} - \dfrac{1}{n_2^2}\right]\)
C. Balmer series
III. \(\bar{v} = R\left[\dfrac{1}{3^2} - \dfrac{1}{n_2^2}\right]\)
D. Brackett series
IV. \(\bar{v} = R\left[\dfrac{1}{4^2} - \dfrac{1}{n_2^2}\right]\)
Choose the correct answer from the options given below:
(A) A-I, B-II, C-III, D-IV
(B) A-IV, B-III, C-II, D-I
(C) A-I, B-III, C-IV, D-II
(D) A-I, B-III, C-II, D-IV
Correct Answer: (D) A-I, B-III, C-II, D-IV
View SolutionStep 1: Understanding the Question:
Each spectral series of hydrogen is a set of lines produced when the electron falls to one fixed lower level. We must match each series name to its wave number formula.
Step 2: Key Formula or Approach:
The general formula is \[ \bar{v} = R\left[\frac{1}{n_1^2} - \frac{1}{n_2^2}\right], \quad n_2 > n_1 \] The number under the first fraction, \(n_1\), is the level where the electron lands. Each series has its own \(n_1\).
Step 3: Fix n1 for Each Series:
Lyman series: \(n_1 = 1\), so formula I. Balmer series: \(n_1 = 2\), so formula II. Paschen series: \(n_1 = 3\), so formula III. Brackett series: \(n_1 = 4\), so formula IV.
Step 4: Write the Matches:
A (Lyman) goes with I. B (Paschen) goes with III. C (Balmer) goes with II. D (Brackett) goes with IV. So the matching is A-I, B-III, C-II, D-IV.
Step 5: Checking Each Option:
Option 1 pairs B with II and C with III. Paschen and Balmer are swapped. Wrong. Option 2 pairs A with IV and D with I. Lyman and Brackett are swapped. Wrong. Option 3 pairs C with IV and D with II. Wrong. Option 4 matches all four pairs. Correct.
Final Answer:
The correct matching is A-I, B-III, C-II, D-IV, which is option 4.
\[ \boxed{4} \]
Question 42:
In Bohr's model of hydrogen atom, which of the following is an integral multiple of \(\dfrac{h}{2\pi}\) ?
(A) Kinetic energy of electron
(B) Radius of an atom
(C) Potential energy of electron
(D) Angular momentum of electron
Correct Answer: (D) Angular momentum of electron
View SolutionStep 1: Understanding the Question:
Bohr's model has one special rule that limits the electron to certain orbits. We must find which quantity is a whole-number multiple of \(h/2\pi\).
Step 2: Key Formula or Approach:
Bohr's quantisation condition says the angular momentum of the electron in an allowed orbit is \[ L = m v r = \frac{n h}{2\pi}, \quad n = 1, 2, 3, \ldots \]
Step 3: Apply the Condition:
The right side is n times \(h/2\pi\), with n a whole number. So the angular momentum is an integral multiple of \(h/2\pi\).
Step 4: Check Each Option:
Option 1: kinetic energy is \(13.6/n^2\) eV. It depends on \(1/n^2\), not on n times \(h/2\pi\). Wrong. Option 2: the radius is \(r_n \propto n^2\). It is measured in metres, not in units of \(h/2\pi\). Wrong. Option 3: potential energy is \(-27.2/n^2\) eV. Wrong. Option 4: angular momentum is \(nh/2\pi\). Correct.
Final Answer:
Only the angular momentum of the electron is quantised as \(nh/2\pi\), so option 4 is correct.
\[ \boxed{\text{Angular momentum of electron}} \]
Question 43:
Arrange the following elements in the increasing order of their binding energies per nucleon. A. \(^{4}\text{He}\) B. \(^{6}\text{Li}\) C. \(^{14}\text{N}\) D. \(^{12}\text{C}\)
Choose the correct answer from the options given below:
(A) A, B, C, D
(B) A, B, D, C
(C) B, A, C, D
(D) B, A, D, C
Correct Answer: (C) B, A, C, D
View SolutionStep 1: Understanding the Question:
Binding energy per nucleon shows how tightly a nucleus is held together. We must list the four nuclei from the lowest value to the highest.
Step 2: Key Approach:
We use the known binding energy per nucleon values from the binding energy curve. He-4: about 7.07 MeV. Li-6: about 5.33 MeV. N-14: about 7.48 MeV. C-12: about 7.68 MeV.
Step 3: Arrange:
Lowest is Li-6 (B), then He-4 (A), then N-14 (C), then C-12 (D). So the increasing order is B, A, C, D.
Step 4: Why These Values Make Sense:
Helium-4 is unusually stable for its size, so it sits above Li-6, which is a loosely bound light nucleus. C-12 is a very stable nucleus and N-14 is slightly lower. For light nuclei the curve is not smooth, so these values must be remembered.
Step 5: Checking Each Option:
Option 1 (A, B, C, D) puts He below Li, which is wrong. Option 2 (A, B, D, C) has the same mistake and also swaps C and D. Option 3 (B, A, C, D) matches our order. Correct. Option 4 (B, A, D, C) swaps N-14 and C-12. Wrong.
Final Answer:
The increasing order of binding energy per nucleon is Li-6, He-4, N-14, C-12, which is option 3.
\[ \boxed{B, A, C, D} \]
Question 44:
If \(r_1\) and \(r_2\) are the radii of the atomic nuclei of mass number 4 and 32, respectively, then the ratio \(\left(\dfrac{r_1}{r_2}\right)\) is
(A) 1:2
(B) 1:3
(C) 1:4
(D) 1:5
Correct Answer: (A) 1:2
View SolutionStep 1: Understand the concept:
The radius of a nucleus depends on its mass number \(A\). Experiments show that nuclear density is nearly the same for all nuclei. So the volume of a nucleus is proportional to the number of nucleons in it.
Step 2: Write the formula:
Since volume is proportional to \(A\) and volume goes as \(R^3\), we get \(R^3 \propto A\). So \[ R = R_0 A^{1/3} \] where \(R_0\) is a constant of about 1.2 fm.
Step 3: Apply it to both nuclei:
For \(A_1 = 4\): \(r_1 = R_0 (4)^{1/3}\). For \(A_2 = 32\): \(r_2 = R_0 (32)^{1/3}\). Dividing, the constant \(R_0\) cancels. \[ \frac{r_1}{r_2} = \left(\frac{4}{32}\right)^{1/3} = \left(\frac{1}{8}\right)^{1/3} \]
Step 4: Simplify:
The cube root of \(\frac{1}{8}\) is \(\frac{1}{2}\). So \(r_1 : r_2 = 1 : 2\).
Step 5: Check the other options:
Options 1:3, 1:4 and 1:5 would need the mass number ratio to be 27, 64 and 125 times, not 8 times. The mass numbers here differ by a factor of 8, so only 1:2 fits.
Final Answer:
The ratio of the radii is 1:2, which is option (1). \[ \boxed{1:2} \]
Question 45:
The maximum wavelength of electromagnetic radiation which can create a hole-electron pair in the semiconductor Ge is: (Given, the band gap of Ge is 0.72 eV)
(A) \(\sim 1.1 \times 10^{-5}\) m
(B) \(\sim 1.4 \times 10^{-5}\) m
(C) \(\sim 1.7 \times 10^{-6}\) m
(D) \(\sim 2.6 \times 10^{-6}\) m
Correct Answer: (C) \(\sim 1.7 \times 10^{-6}\) m
View SolutionStep 1: Understand the concept:
A hole-electron pair forms when a photon lifts an electron from the valence band to the conduction band. The photon energy must be at least the band gap \(E_g\). Larger wavelength means lower energy, so the maximum wavelength matches photon energy equal to \(E_g\).
Step 2: Write the formula:
Photon energy is \(E = \dfrac{hc}{\lambda}\). At the limit, \(E = E_g\), so \[ \lambda_{max} = \frac{hc}{E_g} \] A handy value is \(hc = 1240\) eV nm.
Step 3: Put in the numbers:
\[ \lambda_{max} = \frac{1240 \text{ eV nm}}{0.72 \text{ eV}} \approx 1722 \text{ nm} \] Now convert: \(1722\) nm \(= 1.722 \times 10^{-6}\) m.
Step 4: Match with the options:
The value \(1.72 \times 10^{-6}\) m is close to \(1.7 \times 10^{-6}\) m, which is option (3). Options (1) and (2) are about ten times too large. Option (4) is too large by about 50 percent, so it would need a smaller band gap of about 0.48 eV.
Final Answer:
The maximum wavelength is about \(1.7 \times 10^{-6}\) m, option (3). \[ \boxed{\sim 1.7 \times 10^{-6}\ \text{m}} \]
Question 46:
If the forward bias voltage in a diode is increased, the width of the depletion region
(A) increases
(B) decreases
(C) fluctuates
(D) does not change
Correct Answer: (B) decreases
View SolutionStep 1: Understand the concept:
A p-n junction has a depletion region near the junction. It has no free carriers and only fixed ions. The junction has a built-in potential barrier across it.
Step 2: What forward bias does:
In forward bias the p side is joined to the positive terminal and the n side to the negative terminal. The applied voltage opposes the built-in barrier. So the barrier height drops.
Step 3: Effect on the width:
Free holes and electrons are pushed towards the junction. They neutralise some of the fixed ions in the depletion region. So the depletion region becomes narrower. A larger forward voltage shrinks it further.
Step 4: Check the options:
Option (1) is true for reverse bias, not forward bias. Option (3) is not correct because the width changes smoothly with the voltage. Option (4) is wrong because the width does depend on the bias.
Final Answer:
The depletion region becomes narrower, so the answer is option (2), decreases.
Question 47:
For a heavily doped n-type semi-conductor, donor level lies
(A) a little below the conduction band
(B) a little above the valance band
(C) a little inside the valence band
(D) at the center of the band gap
Correct Answer: (A) a little below the conduction band
View SolutionStep 1: Understand the concept:
In an n-type semiconductor, a pentavalent impurity such as phosphorus is added to silicon. Four of its five valence electrons form bonds. The fifth electron is loosely held.
Step 2: Where the donor level sits:
The energy needed to free this fifth electron is very small, about 0.01 to 0.05 eV. So the donor energy level lies just below the bottom of the conduction band. At room temperature these electrons jump easily into the conduction band.
Step 3: Check option (2) and (3):
A level a little above the valence band, or inside it, is the acceptor level of a p-type semiconductor. So both are wrong here.
Step 4: Check option (4):
A level at the centre of the band gap belongs to an intrinsic semiconductor's Fermi level. It is not a donor level, and electrons there would need a lot of energy to reach the conduction band.
Final Answer:
The donor level lies a little below the conduction band, option (1).
Question 48:
Match List-I with List-II. Assume that the Si intrinsic semiconductor has \(5 \times 10^{28}\) atoms/m\(^3\). (Given: the intrinsic concentration of electrons in the semiconductor is \(10^{16}\ \text{m}^{-3}\)).
LIST-I
LIST-II
A
When it is doped with 2 ppm concentration of a pentavalent atom, \(n_e\) is
I
\(4.0 \times 10^{8}\ \text{m}^{-3}\)
B
When it is doped with 2 ppm concentration of a pentavalent atom, \(n_h\) is
II
\(2.5 \times 10^{23}\ \text{m}^{-3}\)
C
When it is doped with 5 ppm concentration of a pentavalent atom, \(n_e\) is
III
\(10^{9}\ \text{m}^{-3}\)
D
When it is doped with 5 ppm concentration of a pentavalent atom, \(n_h\) is
IV
\(10^{23}\ \text{m}^{-3}\)
Choose the correct answer from the options given below:
(A) A-I, B-II, C-III, D-IV
(B) A-II, B-III, C-IV, D-I
(C) A-IV, B-III, C-II, D-I
(D) A-IV, B-II, C-III, D-I
Correct Answer: (C) A-IV, B-III, C-II, D-I
View SolutionStep 1: Understand the concept:
When a pentavalent atom is added to silicon, each dopant atom gives one free electron. So the electron concentration \(n_e\) is almost equal to the dopant concentration. The hole concentration then follows from the mass action law \(n_e n_h = n_i^2\).
Step 2: Convert ppm to a number density:
1 ppm means 1 dopant atom in \(10^6\) atoms. The number of Si atoms is \(5 \times 10^{28}\) per cubic metre. For 2 ppm: \[ N_D = 2 \times 10^{-6} \times 5 \times 10^{28} = 10^{23}\ \text{m}^{-3} \] For 5 ppm: \[ N_D = 5 \times 10^{-6} \times 5 \times 10^{28} = 2.5 \times 10^{23}\ \text{m}^{-3} \]
Step 3: Find n_e for A and C:
Doping adds far more electrons than the intrinsic \(10^{16}\), so \(n_e \approx N_D\). A (2 ppm): \(n_e = 10^{23}\ \text{m}^{-3}\), which is IV. C (5 ppm): \(n_e = 2.5 \times 10^{23}\ \text{m}^{-3}\), which is II.
Step 4: Find n_h for B and D:
Use \(n_h = \dfrac{n_i^2}{n_e}\) with \(n_i^2 = 10^{32}\). B (2 ppm): \[ n_h = \frac{10^{32}}{10^{23}} = 10^{9}\ \text{m}^{-3} \] which is III. D (5 ppm): \[ n_h = \frac{10^{32}}{2.5 \times 10^{23}} = 4.0 \times 10^{8}\ \text{m}^{-3} \] which is I.
Step 5: Match and check the options:
So A-IV, B-III, C-II, D-I. Option (1) and (2) assign the wrong values to A. Option (4) gives B as II, but \(n_h\) cannot be as high as \(2.5 \times 10^{23}\) in an n-type sample. Only option (3) matches.
Final Answer:
The correct matching is A-IV, B-III, C-II, D-I, which is option (3). \[ \boxed{\text{A-IV, B-III, C-II, D-I}} \]
Question 49:
If 200 MeV energy is released in the fission of a single nucleus of \({}^{235}_{92}U\), how many fissions per second must occur to produce a power of 1 kW?
(A) \(6.250 \times 10^{15}\)
(B) \(3.125 \times 10^{13}\)
(C) \(4.251 \times 10^{13}\)
(D) \(2.155 \times 10^{14}\)
Correct Answer: (B) \(3.125 \times 10^{13}\)
View SolutionStep 1: Understand the concept:
Power is energy released per second. If each fission gives a fixed energy, then the number of fissions per second is total power divided by the energy of one fission.
Step 2: Convert the energy to joules:
Energy of one fission \(= 200\) MeV. Use \(1\text{ MeV} = 1.6 \times 10^{-13}\) J. \[ E = 200 \times 1.6 \times 10^{-13} = 3.2 \times 10^{-11}\ \text{J} \]
Step 3: Convert the power:
Power \(P = 1\) kW \(= 1000\) W \(= 1000\) J/s.
Step 4: Find the fission rate:
\[ n = \frac{P}{E} = \frac{1000}{3.2 \times 10^{-11}} = 3.125 \times 10^{13}\ \text{s}^{-1} \]
Step 5: Check the other options:
Option (1) \(6.25 \times 10^{15}\) is about 200 times too large, which is what you get if the energy is taken as 1 MeV per fission. Options (3) and (4) do not match the calculation. Only option (2) matches.
Final Answer:
About \(3.125 \times 10^{13}\) fissions must occur every second, option (2). \[ \boxed{3.125 \times 10^{13}\ \text{fissions/s}} \]
Question 50:
An alpha particle of energy \(\dfrac{1}{2}\) mv\(^2\) bombards a heavy target nucleus of charge Ze. Then, the distance of closet approach for the alpha particle will be proportional to
(A) \(v^{-1}\)
(B) \(v^{-2}\)
(C) \(v^{-4}\)
(D) \((Ze)^{-1}\)
Correct Answer: (B) \(v^{-2}\)
View SolutionStep 1: Understand the concept:
In head-on approach, the alpha particle slows down because the nucleus repels it. At the closest point, its speed becomes zero and all its kinetic energy has changed into electric potential energy.
Step 2: Write the energy balance:
The alpha particle has charge \(2e\). The potential energy at distance \(r\) from a nucleus of charge \(Ze\) is \(\dfrac{1}{4\pi\epsilon_0}\dfrac{(2e)(Ze)}{r}\). So \[ \frac{1}{2} m v^2 = \frac{1}{4\pi\epsilon_0}\frac{2Ze^2}{r_0} \]
Step 3: Solve for r0:
\[ r_0 = \frac{1}{4\pi\epsilon_0}\frac{2Ze^2}{\frac{1}{2}mv^2} = \frac{Ze^2}{\pi\epsilon_0 m v^2} \] So \(r_0 \propto \dfrac{Ze^2}{m v^2}\).
Step 4: Read the proportionality:
The distance goes as \(v^{-2}\) and grows directly with \(Z\). Option (1) is wrong because the dependence on speed is squared. Option (3) has too high a power. Option (4) is wrong because \(r_0\) is proportional to \(Ze\), not inversely proportional to it.
Final Answer:
The distance of closest approach is proportional to \(v^{-2}\), option (2). \[ \boxed{r_0 \propto v^{-2}} \]
If 200 MeV energy is released in the fission of a single nucleus of \({}^{235}_{92}U\), how many fissions per second must occur to produce a power of 1 kW?
A monochromatic source emitting light of 600 nm, has a power output of 66 W. The number of photons emitted by the source per second is (Given, Plank's constant h = 6.6 \(\times\) 10\(^{-34}\) Js)
A ray of light of frequency \(5 \times 10^{14}\) Hz is passed through a liquid. The wavelength of light measured inside the liquid is found to be \(450 \times 10^{-9}\) m. The refractive index of the liquid is
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