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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2016 (Offline)

MCQ (Single Correct Answer)
A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?
A
0.70 mm
B
0.50 mm
C
0.75 mm
D
0.80 mm

Explanation

Least count = $${{0.5} \over {50}}$$ = 0.01 mm

Zero error = (45 - 50)$$ \times $$0.01 mm = - 0.05 mm

Thickness of sheet = (0.5 + 25$$ \times $$0.01) - (-0.05)

= 0.50 + 0.30 = 0.80 mm
2

JEE Main 2016 (Offline)

MCQ (Single Correct Answer)
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
3

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
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 %
4

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
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

Questions Asked from Units & Measurements

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