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1

### IIT-JEE 2010 Paper 1 Offline

Consider a thin square sheet of side L and thickness, made of a material of resistivity $$\rho$$. The resistance between two opposite faces, shown by the shaded areas in the figure is

A
directly proportional to L.
B
directly proportional to t.
C
independent to L.
D
independent of t.

## Explanation

The sheet is of square shape with thickness t, width $$\omega$$ = L, and length l = L. The resistance between the two opposite faces is given by

$$R = {{\rho l} \over A} = {{\rho l} \over {wt}} = {{\rho L} \over {Lt}} = {\rho \over t}$$.

2

### IIT-JEE 2010 Paper 1 Offline

To verify Ohm's law, a student is provided with a test resitor RT, a high resistance R1, a small resistance R2, two identical galvanometers G1 and G2, and a variable voltage source V. The correct circuit to carry out the experiment is

A
B
C
D

## Explanation

To verify Ohm's law, we need to measure the voltage across the test resistance RT and current passing through it. The voltage can be measured by connecting a high resistance R1 in series with galvanometer. This combination becomes a voltmeter and should be connected in parallel to RT. The current can be measured by connecting a low resistance R2 (shunt) in parallel with galvanometer. This combination becomes an ammeter and should be connected in series to measure the current through RT.

3

### IIT-JEE 2010 Paper 1 Offline

Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100, 60 and 40 W bulbs have filament resistances R100, R60 and R40 respectively, the relation between these resistances is

A
$${1 \over {{R_{100}}}} = {1 \over {{R_{40}}}} + {1 \over {{R_{60}}}}$$
B
$${R_{100}} = {R_{40}} + {R_{60}}$$
C
$${R_{100}} > {R_{60}} > {R_{40}}$$
D
$${1 \over {{R_{100}}}} > {1 \over {{R_{60}}}} > {1 \over {{R_{40}}}}$$

## Explanation

The power of the bulb is

$$P = {{{V^2}} \over R}$$

Therefore,

$$100 = {{{V^2}} \over {{R_{100}}}} \Rightarrow {1 \over {{R_{100}}}} = {{100} \over {{V^2}}}$$

where R100 is the resistance (at any temperature) corresponds to 100 W. Similarly,

$$60 = {{{V^2}} \over {{R_{60}}}} \Rightarrow {1 \over {{R_{60}}}} = {{60} \over {{V^2}}}$$ and $$40 = {{{V^2}} \over {{R_{40}}}} \Rightarrow {1 \over {{R_{40}}}} = {{40} \over {{V^2}}}$$

From these equations, we get

$${P_{100}} > {P_{60}} > {P_{40}} \Rightarrow {1 \over {{R_{100}}}} > {1 \over {{R_{60}}}} > {1 \over {{R_{40}}}}$$

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