MHT CET · Physics · Capacitance

The amount of work done in increasing the voltage across the plates of the capacitor from $5 V$ to $10 V$ is $W$ . The work done in increasing it from $10 V$ to $15 V$ will be

A $W$
B $0.6 W$
C $1.25 W$
D $1.67 W$

Verified Solution

Answer & Solution

Correct Answer

(D) $1.67 W$

Step-by-step Solution

Detailed explanation

The work done in increasing the potential is given by
$W=d U \propto\left(V_2^2-V_1^2\right)$
$\therefore \frac{W_1}{W_2}=\frac{\left(V_2^2-V_1^2\right)}{\left(V_3^2-V_2^2\right)}$ here $V_1=5 v, V_2=10 V$ and $V_3=15 V$
$\therefore \frac{W}{W_2}+\left(\frac{100-25}{225-100}\right)=\frac{75}{125}$
$\therefore W_2=1.67 W$

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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The amount of work done in increasing the voltage across the plates of the capacitor from 5 V to 10 V is W. The work done in increasing it from 10 V to 15 V will be

Step-by-step Solution

The amount of work done in increasing the voltage across the plates of the capacitor from $5 V$ to $10 V$ is $W$ . The work done in increasing it from $10 V$ to $15 V$ will be