MHT CET · Physics · Capacitance
The amount of work done in increasing the voltage across the plates of a capacitor form 5 V to 10 V is ' W '. The work done in increasing it from 10 V to 15 V will be (nearly)
- A 0.6 W
- B W
- C 1.25 W
- D 1.67 W
Answer & Solution
Correct Answer
(D) 1.67 W
Step-by-step Solution
Detailed explanation
\(\mathrm{W}=\frac{1}{2} \mathrm{CV}^2\)
For constant \(\mathrm{C}, \mathrm{W} \propto \mathrm{V}^2\)
\(\begin{array}{ll}
\therefore & \frac{\mathrm{W}_1}{\mathrm{~W}_2}=\frac{\mathrm{V}_2^2-\mathrm{V}_1^2}{\mathrm{~V}_3^2-\mathrm{V}_2^2} \\
& \text { here } \mathrm{V}_1=5 \mathrm{~V}, \mathrm{~V}_2=10 \mathrm{~V}, \mathrm{~V}_3=15 \mathrm{~V} \\
\therefore \quad & \frac{\mathrm{~W}}{\mathrm{~W}_2}=\frac{100-25}{225-100}=\frac{75}{125} \\
\therefore \quad & \mathrm{~W}_2=1.67 \mathrm{~W}
\end{array}\)
For constant \(\mathrm{C}, \mathrm{W} \propto \mathrm{V}^2\)
\(\begin{array}{ll}
\therefore & \frac{\mathrm{W}_1}{\mathrm{~W}_2}=\frac{\mathrm{V}_2^2-\mathrm{V}_1^2}{\mathrm{~V}_3^2-\mathrm{V}_2^2} \\
& \text { here } \mathrm{V}_1=5 \mathrm{~V}, \mathrm{~V}_2=10 \mathrm{~V}, \mathrm{~V}_3=15 \mathrm{~V} \\
\therefore \quad & \frac{\mathrm{~W}}{\mathrm{~W}_2}=\frac{100-25}{225-100}=\frac{75}{125} \\
\therefore \quad & \mathrm{~W}_2=1.67 \mathrm{~W}
\end{array}\)
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