Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

Numerical

The diameter of a spherical bob is measured using a vernier callipers. 9 divisions of the main scale, in the vernier callipers, are equal to 10 divisions of vernier scale. One main scale division is 1 mm. The main scale reading is 10 mm and 8^{th} division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier callipers has positive zero error of 0.04 cm, then the radius of the bob is ___________ $$\times$$ 10^{$$-$$2} cm.

Your Input ________

Correct Answer is **52**

9 MSD = 10 VSD

9 $$\times$$ 1 mm = 10 VSD

$$\therefore$$ 1 VSD = 0.9 mm

LC = 1 MSD $$-$$ 1 VSD = 0.1 mm

Reading = MSR + VSR $$\times$$ LC

10 + 8 $$\times$$ 0.1 = 10.8 mm

Actual reading = 10.8 $$-$$ 0.4 = 10.4 mm

radius = $${d \over 2} = {{10.4} \over 2}$$ = 5.2 mm

= 52 $$\times$$ 10^{$$-$$2} cm

9 $$\times$$ 1 mm = 10 VSD

$$\therefore$$ 1 VSD = 0.9 mm

LC = 1 MSD $$-$$ 1 VSD = 0.1 mm

Reading = MSR + VSR $$\times$$ LC

10 + 8 $$\times$$ 0.1 = 10.8 mm

Actual reading = 10.8 $$-$$ 0.4 = 10.4 mm

radius = $${d \over 2} = {{10.4} \over 2}$$ = 5.2 mm

= 52 $$\times$$ 10

2

Numerical

The acceleration due to gravity is found upto an accuracy of 4% on a planet. The energy supplied to a simple pendulum to known mass 'm' to undertake oscillations of time period T is being estimated. If time period is measured to an accuracy of 3%, the accuracy to which E is known as ..............%

Your Input ________

Correct Answer is **14**

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

$$E = mgl{{{\theta ^2}} \over 2} = m{g^2}{{{T^2}{\theta ^2}} \over {8{\pi ^2}}}$$

$${{dE} \over E} = 2\left( {{{dg} \over g} + {{dT} \over T}} \right)$$

$$ = (4 + 3) = 14\% $$

$$E = mgl{{{\theta ^2}} \over 2} = m{g^2}{{{T^2}{\theta ^2}} \over {8{\pi ^2}}}$$

$${{dE} \over E} = 2\left( {{{dg} \over g} + {{dT} \over T}} \right)$$

$$ = (4 + 3) = 14\% $$

3

Numerical

A 16 $$\Omega$$ wire is bend to form a square loop. A 9V supply having internal resistance of 1$$\Omega$$ is connected across one of its sides. The potential drop across the diagonals of the square loop is _______________ $$\times$$ 10^{$$-$$1} V

Your Input ________

Correct Answer is **45**

Here assume current as

By KVL in outer loop

9 $$-$$ 12i $$-$$ 4i = 0

16i = 9

8i = $${9 \over 2}$$ = 4.5

= 45 $$\times$$ 10^{-1}

By KVL in outer loop

9 $$-$$ 12i $$-$$ 4i = 0

16i = 9

8i = $${9 \over 2}$$ = 4.5

= 45 $$\times$$ 10

4

Numerical

Student A and student B used two screw gauges of equal pitch and 100 equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is 0.322 cm. The absolute value of the difference between the final circular scale readings observed by the students A and B is ______________.

[Figure shows position of reference 'O' when jaws of screw gauge are closed]

Given pitch = 0.1 cm.

[Figure shows position of reference 'O' when jaws of screw gauge are closed]

Given pitch = 0.1 cm.

Your Input ________

Correct Answer is **13**

For (A)

Reading = MSR + CSR + Error

0.322 = 0.300 + CSR + 5 $$\times$$ LC

0.322 = 0.300 + CSR + 0.005

CSR = 0.017

For (B)

Reading = MSR + CSR + Error

0.322 = 0.200 + CSR + 0.092

CSR = 0.030

Difference = 0.030 $$-$$ 0.017 = 0.013 cm

Division on circular scale = $${{0.013} \over {0.001}} = 13$$

Reading = MSR + CSR + Error

0.322 = 0.300 + CSR + 5 $$\times$$ LC

0.322 = 0.300 + CSR + 0.005

CSR = 0.017

For (B)

Reading = MSR + CSR + Error

0.322 = 0.200 + CSR + 0.092

CSR = 0.030

Difference = 0.030 $$-$$ 0.017 = 0.013 cm

Division on circular scale = $${{0.013} \over {0.001}} = 13$$

On those following papers in Numerical

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JEE Main 2021 (Online) 31st August Evening Shift (1)

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JEE Main 2021 (Online) 25th July Morning Shift (1)

JEE Main 2021 (Online) 22th July Evening Shift (1)

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JEE Main 2020 (Online) 6th September Morning Slot (1)

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