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### JEE Main 2021 (Online) 27th August Morning Shift

Numerical
Two persons A and B perform same amount of work in moving a body through a certain distance d with application of forces acting at angle 45$$^\circ$$ and 60$$^\circ$$ with the direction of displacement respectively. The ratio of force applied by person A to the force applied by person B is $${1 \over {\sqrt x }}$$. The value of x is .................... .

## Explanation

Given WA = WB

FAd cos45$$^\circ$$ = FBd cos60$$^\circ$$

$${F_A} \times {1 \over {\sqrt 2 }} = {F_B} \times {1 \over 2}$$

$${{{F_A}} \over {{F_B}}} = {{\sqrt 2 } \over 2} = {1 \over {\sqrt 2 }}$$

x = 2
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### JEE Main 2021 (Online) 26th August Morning Shift

Numerical
A uniform chain of length 3 meter and mass 3 kg overhangs a smooth table with 2 meter lying on the table. If k is the kinetic energy of the chain in joule as it completely slips off the table, then the value of k is ................. . (Take g = 10 m/s2)

## Explanation From energy conservation

Ki + Ui = kf + Uf

$$0 + \left( { - 1 \times 10 \times {1 \over 2}} \right) = {k_f} + \left( { - 3 \times 10 \times {3 \over 2}} \right)$$

$$-$$5 = kf $$-$$ 45

kf = 40 J
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### JEE Main 2021 (Online) 27th July Evening Shift

Numerical
A small block slides down from the top of hemisphere of radius R = 3 m as shown in the figure. The height 'h' at which the block will lose contact with the surface of the sphere is __________ m.

(Assume there is no friction between the block and the hemisphere) ## Explanation $$mg\cos \theta = {{m{v^2}} \over R}$$ .... (1)

$$\cos \theta = {h \over R}$$

Energy conservation

$$mg\{ R - h\} = {1 \over 2}m{v^2}$$ ..... (2)

from (1) & (2)

$$\Rightarrow$$ $$mg\left\{ {{h \over R}} \right\} = {{2mg\{ R - h\} } \over R}$$

$$h = {{2R} \over 3}$$ = 2m
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### JEE Main 2021 (Online) 25th July Evening Shift

Numerical
A force of F = (5y + 20)$$\widehat j$$ N acts on a particle. The work done by this force when the particle is moved from y = 0 m to y = 10 m is ___________ J.

## Explanation

F = (5y + 20)$$\widehat j$$

$$W = \int {Fdy = \int\limits_0^{10} {(5y + 20)dy} }$$

$$= \left( {{{5{y^2}} \over 2} + 20y} \right)_0^{10}$$

$$= {5 \over 2} \times 100 + 20 \times 10$$

$$= 250 + 200 = 450$$ J

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