### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2015 (Offline)

The period of oscillation of a simple pendulum is $T = 2\pi \sqrt {{L \over g}}$. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:
A
1 %
B
5 %
C
2 %
D
3 %

## Explanation

Given $T = 2\pi \sqrt {{L \over g}}$

$\Rightarrow g = {{4{\pi ^2}L} \over {{T^2}}}$

$\Rightarrow g = {{4{\pi ^2}L{n^2}} \over {{t^2}}}$

[ as $T = {t \over n}$ ]

So, percentage error in $g$ =

${{\Delta g} \over g} \times 100 = {{\Delta L} \over L} \times 100 + 2{{\Delta t} \over t} \times 100$

= ${{0.1} \over {20.0}} \times 100 + 2 \times {1 \over {90}} \times 100$

= 2.72 % = 3 %
2

### JEE Main 2014 (Offline)

A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?
A
A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.
B
A screw gauge having 50 divisions in the circular scale and pitch as 1 mm.
C
A meter scale.
D
A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm.

## Explanation

Measured length of rod = 3.50 cm

That means least count of the measuring instrument should be 0.01 cm = 0.1 mm

For vernier scale 1 main scale division = 1 mm

And 9 MSD = 10 VSD

Least count = 1 MSD - 1 VSD

= 1 - 0.9 = 0.1 mm
3

### JEE Main 2013 (Offline)

Let [${\varepsilon _0}$] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then:
A
${\varepsilon _0} = \left[ {{M^{ - 1}}{L^{ - 3}}{T^2}A} \right]$
B
${\varepsilon _0} =$$\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$
C
${\varepsilon _0} = \left[ {{M^1}{L^2}{T^1}{A^2}} \right]$
D
${\varepsilon _0} = \left[ {{M^1}{L^2}{T^1}A} \right]$

## Explanation

From Coulomb's law we know,

$F = {1 \over {4\pi { \in _0}}}{{{q_1}{q_2}} \over {{r^2}}}$

$\therefore$ ${ \in _0} = {1 \over {4\pi }}{{{q_1}{q_2}} \over {F{r^2}}}$

Hence, $\left[ {{ \in _0}} \right] = {{\left[ {AT} \right]\left[ {AT} \right]} \over {\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]}}$

= $\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$
4

### AIEEE 2012

Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is
A
6 %
B
zero
C
1 %
D
3 %

## Explanation

We know R = ${V \over I}$

$\therefore$ ${{\Delta R} \over R} = {{\Delta V} \over V} + {{\Delta I} \over I}$

Percentage error in R =

${{\Delta R} \over R} \times 100 = {{\Delta V} \over V} \times 100 + {{\Delta I} \over I} \times 100$

$\therefore$ ${{\Delta R} \over R} \times 100 =$ 3% + 3 % = 6%