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Physics · 2026 · 4 marks
CBSE 2026 · Region 4 · Set 2 · Q30
A student sets up the circuit as shown in the figure to find the value of unknown resistance X and records a set of readings of the voltmeter and the…
A student sets up the circuit as shown in the figure to find the value of unknown resistance X and records a set of readings of the voltmeter and the ammeter by using the rheostat.
(i)
If resistor X were made of manganin and readings for V and I are taken without switching off the circuit, the graph between V and I will be as :
(A)
(B)
(C)
(D)
(ii)
Error in the value of X obtained from different sets of voltmeter and ammeter readings, is :
(A)
due to error in voltmeter reading only.
(B)
due to error in ammeter reading only.
(C)
equal to the sum of error in voltmeter reading and error in ammeter reading.
(D)
equal to error in voltmeter reading divided by the error in ammeter reading.
(iii)
If the movable end of rheostat is moved towards P, then :
(A)
reading in ammeter decreases and reading in voltmeter increases.
(B)
readings in both voltmeter and ammeter increase.
(C)
reading in ammeter increases and reading in voltmeter decreases.
(D)
readings in both voltmeter and ammeter decrease.
(iv)(a)
Suppose the unknown resistance X is replaced by a wire made of the same metal. This wire consists of three parts, of the same length L but has radii r, r/$\displaystyle 3$ and r/$\displaystyle 2$ as shown in the figure.
For a particular setting of the rheostat, let $\displaystyle \mathrm{v}_{1}, \mathrm{v}_{2}$ and $\displaystyle \mathrm{v}_{3}$ be the value of drift velocities in parts AC, CD and DB. Then :
(A)
$\displaystyle \mathrm{v}_{1}>\mathrm{v}_{2}>\mathrm{v}_{3}$
(B)
$\displaystyle \mathrm{v}_{2}>\mathrm{v}_{3}>\mathrm{v}_{1}$
(C)
$\displaystyle \mathrm{v}_{3}>\mathrm{v}_{2}>\mathrm{v}_{1}$
(D)
$\displaystyle \mathrm{v}_{1}=\mathrm{v}_{2}=\mathrm{v}_{3}$
Consider the same wire, as shown in figure in question
(iv)(a)
connected in place of X. For a particular setting of rheostat, let $\displaystyle \mathrm{E}_{1}, \mathrm{E}_{2}$ and $\displaystyle \mathrm{E}_{3}$ be the value of electric fields in part AC, CD and DB. Then :
(A)
$\displaystyle \mathrm{E}_{1}=\mathrm{E}_{2}=\mathrm{E}_{3}$
(B)
$\displaystyle \mathrm{E}_{3}>\mathrm{E}_{2}>\mathrm{E}_{1}$
(C)
$\displaystyle \mathrm{E}_{2}>\mathrm{E}_{3}>\mathrm{E}_{1}$
(D)
$\displaystyle \mathrm{E}_{1}>\mathrm{E}_{2}>\mathrm{E}_{3}$
Marking-scheme solution
(i)(C)
(ii)(C)
equal to the sum of the error in the voltmeter reading and the error in the ammeter reading.
(iii)(B)
readings in both voltmeter and ammeter increase.
(iv)(B)
$\displaystyle \mathrm{v}_{2}>\mathrm{v}_{3}>\mathrm{v}_{1}$ OR (C) $\displaystyle \mathrm{E}_{2}>\mathrm{E}_{3}>\mathrm{E}_{1}$
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