(a) Find the ratio of resistances of two copper rods X and Y of lengths 30 cm and 10 cm respectively and having radii 2 cm and 1 cm respectively.(b) A current of 500 mA flows in a series circuit containing an electric lamp and a conductor of 10W when connected to 6V battery. Find the resistance of the electric lamp.

Academic Physics NCERT Class 10

(a) Given:

Length of wire $X$ = 30 cm

Length of wire $Y$ = 10 cm

Radius of wire $X$ = 2 cm

Radius of wire $Y$ = 1 cm

To find: Ratio of resistance of two copper rods $X$ and $Y$.

Solution:

We know, that-

$R=\frac {ρl}{A}$

where,

$ρ$ = Specific Resistance (it is invariable for the same material)

$A$ = Cross sectional area

$l$ = Length of wire

Therefore,

Specific resistance for the copper rod $X$

$R_X=\frac {ρl_X}{A_X}\Rightarrow \frac {ρ30}{\pi (2^2)}\Rightarrow \frac {ρ30}{4\pi}$ $[Area\ of\ cross\ section,\ A=\pi r^2]$

Specific resistance for the copper rod $Y$

$R_Y=\frac {ρl_Y}{A_Y}\Rightarrow \frac {ρ10}{\pi (1^2)}\Rightarrow \frac {ρ10}{1\pi}$ $[Area\ of\ cross\ section,\ A=\pi r^2]$

Thus, the ratio of the resistance of two copper rods $X$ and $Y$ is-

$\frac {R_X}{R_Y}=\frac {\frac {ρ30}{4\pi}}{\frac {ρ10}{1\pi}}$

$\frac {R_X}{R_Y}=\frac {ρ30}{4\pi}\times {\frac {1\pi}{ρ10}}$

$\frac {R_X}{R_Y}=\frac {30}{40}$

$\frac {R_X}{R_Y}=\frac {3}{4}$

$R_X: R_Y= 3:4$

(b) Given:

Current, $I$ = 500 mA = 0.5 A $ (converted milliampere to ampere)

Resistance of the conductor, $R_C$ = $10\Omega$

Potential difference, $V$ = 6 V

To find: Resistance of lamp, $R_L$.

Solution:

As the lamp, $R_L$ and conductor, $R_C$ are connected in series, then the totak resistance of the circuit is given as-

$R_{T}=R_1+R_2$ or $R_{net}=R_L+R_C$

$R_{T}=R_L+10$

Now, we know that the formula of resistance is given as-

$R=\frac {V}{I}$

Putting the given values we get-

$R_T=\frac {6}{0.5}$

$R_L+10=\frac {60}{5}$

$R_L+10=12$

$R_L=12-10$

$R_L=2\Omega$

Thus, the resistance of the electric lamp is 2 Ohm.

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Updated on: 10-Oct-2022

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