JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 4)
The normal reaction 'N' for a vehicle of 800 kg mass, negotiating a turn on a 30$$^\circ$$ banked road at maximum possible speed without skidding is ____________ $$\times$$ 103 kg m/s2. [Given cos30$$^\circ$$ = 0.87, $$\mu$$s = 0.2]
12.4
7.2
6.96
10.2
Explanation
The given situation can be represented as
_20th_July_Morning_Shift_en_4_1.png)
Equating forces perpendicular to the inclined plane,
$$N = mg\cos 30^\circ + {{m{v^2}} \over R}\sin 30^\circ $$
$$ \Rightarrow N - mg\cos 30^\circ = {{m{v^2}} \over R}\sin 30^\circ $$ .... (i)
Equating forces along the inclined plane,
$$mg\sin 30^\circ + {\mu _s}N = {{m{v^2}} \over R}\cos 30^\circ $$ .... (ii)
On dividing Eq. (i) by Eq. (ii), we get
$${{N - mg\cos 30^\circ } \over {mg\sin 30^\circ + {\mu _s}N}} = \tan 30^\circ $$ [$$\because$$ $$\cos 30^\circ = {{\sqrt 3 } \over 2}$$ and $$\sin 30^\circ = {1 \over 2}$$]
$$\Rightarrow$$ $${{N - mg(\sqrt3/2) } \over {mg(1/2) + (0.2)N}} = {1 \over {\sqrt 3 }}$$
$$N\sqrt 3 - mg{{\sqrt 3 } \over 2}.\sqrt 3 = {{mg} \over 2} + 0.2N$$
$$ \Rightarrow (\sqrt 3 - 0.2)N = mg{{(1 + 3)} \over 2} = 2mg$$
$$ \Rightarrow N = {{2mg} \over {\sqrt 3 - 0.2}} = {{2 \times 800 \times 10} \over {1.532}}$$
$$ = 10.44 \times {10^3}V$$
Therefore, $$N = 10.2 \times {10^3}$$ kg-m/s2
Equating forces perpendicular to the inclined plane,
$$N = mg\cos 30^\circ + {{m{v^2}} \over R}\sin 30^\circ $$
$$ \Rightarrow N - mg\cos 30^\circ = {{m{v^2}} \over R}\sin 30^\circ $$ .... (i)
Equating forces along the inclined plane,
$$mg\sin 30^\circ + {\mu _s}N = {{m{v^2}} \over R}\cos 30^\circ $$ .... (ii)
On dividing Eq. (i) by Eq. (ii), we get
$${{N - mg\cos 30^\circ } \over {mg\sin 30^\circ + {\mu _s}N}} = \tan 30^\circ $$ [$$\because$$ $$\cos 30^\circ = {{\sqrt 3 } \over 2}$$ and $$\sin 30^\circ = {1 \over 2}$$]
$$\Rightarrow$$ $${{N - mg(\sqrt3/2) } \over {mg(1/2) + (0.2)N}} = {1 \over {\sqrt 3 }}$$
$$N\sqrt 3 - mg{{\sqrt 3 } \over 2}.\sqrt 3 = {{mg} \over 2} + 0.2N$$
$$ \Rightarrow (\sqrt 3 - 0.2)N = mg{{(1 + 3)} \over 2} = 2mg$$
$$ \Rightarrow N = {{2mg} \over {\sqrt 3 - 0.2}} = {{2 \times 800 \times 10} \over {1.532}}$$
$$ = 10.44 \times {10^3}V$$
Therefore, $$N = 10.2 \times {10^3}$$ kg-m/s2
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