JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 6)
A block of mass m slides on the wooden wedge, which in turn slides backward on the horizontal surface. The acceleration of the block with respect to the wedge is : Given m = 8 kg, M = 16 kg Assume all the surfaces shown in the figure to be frictionless.
_1st_September_Evening_Shift_en_6_1.png)
$${4 \over 3}g$$
$${6 \over 5}g$$
$${3 \over 5}g$$
$${2 \over 3}g$$
Explanation
Let acceleration of wedge is aw and acceleration of block w.r.t. wedge is ab
For the wedge w.r.t. ground
_1st_September_Evening_Shift_en_6_2.png)
N cos60$$^\circ$$ = Maw = 16aw
$$\Rightarrow$$ N = 32aw
F.B.D. of block w.r.t. wedge
_1st_September_Evening_Shift_en_6_3.png)
Balancing vertical forces
N = 8g cos30$$^\circ$$ $$-$$ 8aw sin30$$^\circ$$ $$\Rightarrow$$ 32aw = 4$$\sqrt 3 $$g $$-$$ 4aw
$$\Rightarrow$$ aw = $${{\sqrt 3 } \over 9}$$g
Along incline plane
8g sin30$$^\circ$$ + 8aw cos30$$^\circ$$ = mab = 8ab
ab = g $$\times$$ $${1 \over 2} + {{\sqrt 3 } \over 9}$$g . $${{\sqrt 3 } \over 2} = {{2g} \over 3}$$
Option (d)
For the wedge w.r.t. ground
N cos60$$^\circ$$ = Maw = 16aw
$$\Rightarrow$$ N = 32aw
F.B.D. of block w.r.t. wedge
Balancing vertical forces
N = 8g cos30$$^\circ$$ $$-$$ 8aw sin30$$^\circ$$ $$\Rightarrow$$ 32aw = 4$$\sqrt 3 $$g $$-$$ 4aw
$$\Rightarrow$$ aw = $${{\sqrt 3 } \over 9}$$g
Along incline plane
8g sin30$$^\circ$$ + 8aw cos30$$^\circ$$ = mab = 8ab
ab = g $$\times$$ $${1 \over 2} + {{\sqrt 3 } \over 9}$$g . $${{\sqrt 3 } \over 2} = {{2g} \over 3}$$
Option (d)
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