JEE Mains Previous Years Questions with Solutions Android App

Download our App

JEE Mains Previous Years Questions with Solutions

4.5 
Star 1 Star 2 Star 3 Star 4
Star 5
  (100k+ )
1

JEE Main 2021 (Online) 1st September Evening Shift

Numerical
A steel rod with y = 2.0 $$\times$$ 1011 Nm$$-$$2 and $$\alpha$$ = 10$$-$$5 $$^\circ$$C$$-$$1 of length 4 m and area of cross-section 10 cm2 is heated from 0$$^\circ$$C to 400$$^\circ$$C without being allowed to extend. The tension produced in the rod is x $$\times$$ 105 N where the value of x is ____________.
Your Input ________

Answer

Correct Answer is 8

Explanation

Given, the Young's modulus of the steel rod, Y = 2 $$\times$$ 1011 Pa

Thermal coefficient of the steel rod, $$\alpha$$ = 10$$-$$5$$^\circ$$C

The length of the steel rod, l = 4 m

The area of the cross-section, A = 10 cm2

The temperature difference, $$\Delta$$T = 400$$^\circ$$C

As we know that,

Thermal strain = $$\alpha$$ $$\Delta$$T

Using the Hooke's law

Young's modulus (Y) = $${{Thermal\,stress} \over {Thermal\,strain}} = {{F/A} \over {\alpha \,\Delta \,T}}$$

Thermal stress, $$F = YA\,\alpha \,\Delta \,T$$

Substitute the values in the above equation, we get

$$F = 2 \times {10^{11}} \times 10 \times {10^{ - 4}} \times {10^{ - 5}} \times (400)$$

$$ = 8 \times {10^5}N$$

Comparing with, $$F = x \times {10^5}N$$

The value of the x = 8.
2

JEE Main 2021 (Online) 31st August Morning Shift

Numerical
When a rubber ball is taken to a depth of __________m in deep sea, its volume decreases by 0.5%. (The bulk modulus of rubber = 9.8 $$\times$$ 108 Nm$$-$$2, Density of sea water = 103 kgm$$-$$3, g = 9.8 m/s2)
Your Input ________

Answer

Correct Answer is 500

Explanation

$$B = - {{\Delta P} \over {\left( {{{\Delta V} \over V}} \right)}} = - {{\rho gh} \over {\left( {{{\Delta V} \over V}} \right)}}$$

$$ - {{B{{\Delta V} \over V}} \over {\rho g}} = h$$

$${{9.8 \times {{10}^8} \times 0.5} \over {100 \times {{10}^3} \times 9.8}} = h$$

h = 500
3

JEE Main 2021 (Online) 27th August Evening Shift

Numerical
Wires W1 and W2 are made of same material having the breaking stress of 1.25 $$\times$$ 109 N/m2. W1 and W2 have cross-sectional area of 8 $$\times$$ 10$$-$$7 m2 and 4 $$\times$$ 10$$-$$7 m2, respectively. Masses of 20 kg and 10 kg hang from them as shown in the figure. The maximum mass that can be placed in the pan without breaking the wires is ____________ kg. (Use g = 10 m/s2)

Your Input ________

Answer

Correct Answer is 40

Explanation

B.S1 = $${{{T_{1\max }}} \over {8 \times {{10}^{ - 7}}}}$$ $$\Rightarrow$$ T1 max = 8 $$\times$$ 1.25 $$\times$$ 100 = 1000 N

B.S2 = $${{{T_{2\max }}} \over {4 \times {{10}^{ - 7}}}}$$ $$\Rightarrow$$ T2 max = 4 $$\times$$ 1.25 $$\times$$ 100 = 500 N

m = $${{500 - 100} \over {10}} = 40$$ kg
4

JEE Main 2021 (Online) 26th August Morning Shift

Numerical
A soap bubble of radius 3 cm is formed inside the another soap bubble of radius 6 cm. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is ................ cm.
Your Input ________

Answer

Correct Answer is 2

Explanation

Image

Excess pressure inside the smaller soap bubble

$$\Delta P = {{4S} \over {{r_1}}} + {{4S} \over {{r_2}}}$$ .... (i)

The excess pressure inside equivalent soap bubble

$$\Delta P = {{4S} \over {{R_{eq}}}}$$ ....... (ii)

From (i) & (ii)

$${{4S} \over {{R_{eq}}}} = {{4S} \over {{r_1}}} + {{4S} \over {{r_2}}}$$

$${1 \over {{R_{eq}}}} = {1 \over {{r_1}}} + {1 \over {{r_2}}}$$

$$ = {1 \over 6} + {1 \over 3}$$

Req = 2 cm

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12