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1

### IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)

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

MCQ (Single Correct Answer)

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

MCQ (Single Correct Answer)

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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