JEE MAIN - Physics (2019 - 12th January Evening Slot - No. 17)
A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force 2 N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10 N. The coefficient of static friction between the block and the plane is : [Take g = 10 m/s2]
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$${{\sqrt 3 } \over 4}$$
$${1 \over 2}$$
$${{\sqrt 3 } \over 2}$$
$${2 \over 3}$$
Explanation
2 + mg sin30 = $$\mu $$mg cos30o
10 = mgsin30 + $$\mu $$ mg cos30o
= 2$$\mu $$mg cos30 $$-$$ 2
6 = $$\mu $$mg cos30
4 = mg sin30
$${3 \over 2} = \mu \times \sqrt 3 $$
$$\mu $$ = $${{\sqrt 3 } \over 2}$$
10 = mgsin30 + $$\mu $$ mg cos30o
= 2$$\mu $$mg cos30 $$-$$ 2
6 = $$\mu $$mg cos30
4 = mg sin30
$${3 \over 2} = \mu \times \sqrt 3 $$
$$\mu $$ = $${{\sqrt 3 } \over 2}$$
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