A car is travelling along a flyover bridge which is a part of vertical circle of radius $10 m$ . At the highest point of it if the normal reaction on the car is half of its weight, the speed of car is:

Question

A car is travelling along a flyover bridge which is a part of vertical circle of radius

10 m

. At the highest point of it if the normal reaction on the car is half of its weight, the speed of car is:

Toppr Admin · Accepted Answer

Normal reaction is half-xA0- its weight-x21D2-N-mg2Net force which provides centripetal force ismg-x2212-mg2-mv2R-x2192-v-x221A-Rg2-x221A-9-8-xD7-102-7m-s

A car is travelling along a flyover bridge which is a part of vertical circle of radius $10 m$ . At the highest point of it if the normal reaction on the car is half of its weight, the speed of car is:
$7 m / s$
$10 m / s$
$14 m / s$
$20 m / s$

Normal reaction is half its weight
$\Rightarrow N = \frac{m g}{2}$
Net force which provides centripetal force is
$m g - \frac{m g}{2} = \frac{m v^{2}}{R}$
$\to v = \sqrt{\frac{R g}{2}} = \sqrt{\frac{9.8 \times 10}{2}} = 7 m / s$

A car is travelling along a flyover bridge which is a part of vertical circle of radius 10 m. At the highest point of it if the normal reaction on the car is half of its weight, the speed of car is:7 m/s10 m/s14 m/s20 m/s

Normal reaction is half its weight⇒N=mg2Net force which provides centripetal force ismg−mg2=mv2R→v=√Rg2=√9.8×102=7m/s

A car is travelling along a flyover bridge which is a part of vertical circle of radius $10 m$ . At the highest point of it if the normal reaction on the car is half of its weight, the speed of car is:
$7 m / s$
$10 m / s$
$14 m / s$
$20 m / s$

Normal reaction is half its weight
$\Rightarrow N = \frac{m g}{2}$
Net force which provides centripetal force is
$m g - \frac{m g}{2} = \frac{m v^{2}}{R}$
$\to v = \sqrt{\frac{R g}{2}} = \sqrt{\frac{9.8 \times 10}{2}} = 7 m / s$