1
JEE Main 2022 (Online) 28th June Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A water drop of radius 1 $$\mu$$m falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8 $$\times$$ 10$$-$$5 Nsm$$-$$2 and its density is negligible as compared to that of water 106 gm$$-$$3. Terminal velocity of the water drop is :

(Take acceleration due to gravity = 10 ms$$-$$2)

A
145.4 $$\times$$ 10$$-$$6 ms$$-$$1
B
118.0 $$\times$$ 10$$-$$6 ms$$-$$1
C
132.6 $$\times$$ 10$$-$$6 ms$$-$$1
D
123.4 $$\times$$ 10$$-$$6 ms$$-$$1
2
JEE Main 2022 (Online) 28th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : Product of Pressure (P) and time (t) has the same dimension as that of coefficient of viscosity.

Reason R : Coefficient of viscosity = $${{Force} \over {Velocity\,gradient}}$$

Choose the correct answer from the options given below :

A
Both A and R are true, and R is the correct explanation of A.
B
Both A and R are true but R is NOT the correct explanation of A.
C
A is true but R is false.
D
A is false but R is true.
3
JEE Main 2022 (Online) 28th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075 N/m. In this process the gain in surface energy will be :

A
2.8 $$\times$$ 10$$-$$4 J
B
1.5 $$\times$$ 10$$-$$3 J
C
1.9 $$\times$$ 10$$-$$4 J
D
9.4 $$\times$$ 10$$-$$5 J
4
JEE Main 2022 (Online) 27th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The velocity of a small ball of mass 'm' and density d1, when dropped in a container filled with glycerin, becomes constant after some time. If the density of glycerin is d2, then the viscous force acting on the ball, will be :

A
$$mg\left( {1 - {{{d_1}} \over {{d_2}}}} \right)$$
B
$$mg\left( {1 - {{{d_2}} \over {{d_1}}}} \right)$$
C
$$mg\left( {{{{d_1}} \over {{d_2}}} - 1} \right)$$
D
$$mg\left( {{{{d_2}} \over {{d_1}}} - 1} \right)$$
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