CBSE Class 10 Science 2024: question 36(b)

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. (i) The potential difference between two points is 1 volt when 1 joule of work is done to move a charge of 1 coulomb from one point to the other: $1\text{V}=\frac{1\text{J}}{1\text{C}}$.

   1 mark

2. (ii) Given: $d=0.2\text{mm}=2\times 10^{-4}\text{m}$, $\rho =1.6\times 10^{-8}\Omega \text{m}$, $R=14\Omega$.

   Area of cross-section: $A=\frac{\pi d^2}{4}$.

   ½ mark

3. From $R=\rho \frac{l}{A}$ we get $l=\frac{RA}{\rho }=\frac{R\pi d^2}{4\rho }$.

   ½ mark

4. $l=\frac{14\times \frac{22}{7}\times (2\times 10^{-4}{)}^2}{4\times 1.6\times 10^{-8}}=\frac{14\times \frac{22}{7}\times 4\times 10^{-8}}{6.4\times 10^{-8}}=\frac{44}{1.6}=27.5\text{m}$

   1 mark

5. Diameter doubled: $d^{\prime }=2d$, so $A^{\prime }=\frac{\pi (2d{)}^2}{4}=4A$. Since $R\propto \frac{1}{A}$ (same $\rho$ and $l$), $\frac{R^{\prime }}{R}=\frac{A}{A^{\prime }}=\frac{1}{4}$.

   ½ mark

6. $R^{\prime }=\frac{R}{4}=\frac{14}{4}=3.5\Omega$

   1 mark

7. Change in resistance $=14.0-3.5=10.5\Omega$. The resistance decreases by $10.5\Omega$, becoming one-fourth of its original value.

   ½ mark

Answer: (i) When 1 J of work is done to move 1 C of charge between the points. (ii) l = 27.5 m; on doubling the diameter, R becomes 3.5 Ω, a decrease of 10.5 Ω

Revise the chapter

• Chapter: Electricity (notes, formulas, where marks are lost and every past question)
• Topic: Resistance and Resistivity