Resistance in meter bridge’s two arms are 5 ohms and R ohms. When resistance R is shunted with equal resistance, new balance point becomes 1.6l1. Calculate R.

Resistance can be defined as the measurement of opposition of current flow in an electrical circuit. Therefore, in order to determine the value of resistance (R), we need to:

Known resistance, R = 5Ω
Initial Unknown Resistance = RΩ
New Balancing Point, L = 1.6l1

Thus, R/L = S/100 − L … [1]

Here,

R = Known Resistance
S = Unknown Resistance

In the first stage, the balancing point is at l₁.

Now, after replacing the values as:

R = 5Ω and L = l₁ in equation [1], we can obtain,

⇒ 5/l₁ = S/100−l₁ … [2]

Now, after resistance R is shunted with equal resistance R, the resistance in that arm can be R₂

Then,

⇒ 5L = R/2/100−L … [3]

Replace the value of L1 in equation 3. We can obtain,

⇒ 5/1.6l₁ = R/2/100−1.6l1 … [4]

Hence, divide the given equation 2 by 4, in order to get,

1.6/2 = (100 − 1.6l1)/100 − l1

⇒ 160 − 1.6l1 = 200 − 3.2l1

⇒ 1.6l₁= 40

⇒ l₁= 401.6

= 25 cm

Replace l₁=25cm in the above equation 1, we get,

5/25 = R/100 − 25

⇒ R = 15Ω

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Resistance in meter bridge’s two arms are 5 ohms and R ohms. When resistance R is shunted with equal resistance, new balance point becomes 1.6l1. Calculate R.

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A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.

3.
A rectangular loop of size 5 cm $\times$ 8 cm is lying in the $x$-$y$ plane in a uniform magnetic field $\vec{B} = (2.0\ \text{T})\hat{k}$. The total magnetic flux linked with the loop is:

5.
The resistance of a metal wire at \(20^\circ\text{C}\) is \(1.05\,\Omega\) and at \(100^\circ\text{C}\) is \(1.38\,\Omega\). Determine the temperature coefficient of resistivity of this metal.

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Resistance in meter bridge’s two arms are 5 ohms and R ohms. When resistance R is shunted with equal resistance, new balance point becomes 1.6l1. Calculate R.

Related Questions

CBSE CLASS XII Related Questions

2.A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.

3.A rectangular loop of size 5 cm $\times$ 8 cm is lying in the $x$-$y$ plane in a uniform magnetic field $\vec{B} = (2.0\ \text{T})\hat{k}$. The total magnetic flux linked with the loop is:

5.The resistance of a metal wire at \(20^\circ\text{C}\) is \(1.05\,\Omega\) and at \(100^\circ\text{C}\) is \(1.38\,\Omega\). Determine the temperature coefficient of resistivity of this metal.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.

3.
A rectangular loop of size 5 cm $\times$ 8 cm is lying in the $x$-$y$ plane in a uniform magnetic field $\vec{B} = (2.0\ \text{T})\hat{k}$. The total magnetic flux linked with the loop is:

5.
The resistance of a metal wire at \(20^\circ\text{C}\) is \(1.05\,\Omega\) and at \(100^\circ\text{C}\) is \(1.38\,\Omega\). Determine the temperature coefficient of resistivity of this metal.