### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2016 (Offline)

A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be :
A
92 $\pm$ 1.8 s
B
92 $\pm$ 3 s
C
92 $\pm$ 2 s
D
92 $\pm$ 5.0 s

## Explanation

Here t1 = 90 s, t2 = 91 s, t3 = 95 s, t4 = 92 s

Mean(t) = ${{{t_1} + {t_2} + {t_3} + {t_4}} \over 4}$

= ${{90 + 91 + 95 + 92} \over 4}$ = 92 s

Now mean deviation

= ${{2 + 1 + 3 + 0} \over 4}$ = 1.5 s

Since least count of clock is one second, so reported mean time

= (92 $\pm$ 2) s
2

### 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 %
3

### 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
4

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

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