A smooth inclined plane is inclined at an angle θ with the horizontal. A body starts from rest and slides down the inclined surface, then the time taken by the body to reach the bottom is

Question

A smooth inclined plane is inclined at an angle θ with the horizontal. A body starts from rest and slides down the inclined surface, then the time taken by the body to reach the bottom is (A) √(2 h/g) (B) √(21/g) (C) (1/ sin θ) √(2 h/g) (D)  sin θ √(2 h/g)

Tardigrade Admin · Accepted Answer

Acceleration due to gravity along inclined plane

g^{1} = g cos (9 0^{\circ} - θ) = g sin θ

∴

Time taken,

t = \frac{2 s}{g ^{'}} = \frac{2 ℓ}{g s i n θ}

But

sin θ = \frac{h}{ℓ} \Rightarrow ℓ = \frac{h}{s i n θ}

Hence,

t = \frac{2}{g s i n θ} - \frac{h}{s i n θ}

t = \frac{1}{s i n θ} \frac{2 h}{g}

Q. A smooth inclined plane is inclined at an angle $θ$ with the horizontal. A body starts from rest and slides down the inclined surface, then the time taken by the body to reach the bottom is

$\frac{2 h}{g}$

$\frac{21}{g}$

$\frac{1}{s i n θ} \frac{2 h}{g}$

$sin θ \frac{2 h}{g}$

Q. A smooth inclined plane is inclined at an angle θ with the horizontal. A body starts from rest and slides down the inclined surface, then the time taken by the body to reach the bottom is

g2h​​

g21​​

sinθ1​g2h​​

sinθg2h​​

Acceleration due to gravity along inclined plane g1=gcos(90∘−θ)=gsinθ ∴ Time taken, t=g′2s​​=gsinθ2ℓ​​ But sinθ=ℓh​⇒ℓ=sinθh​ Hence, t=gsinθ2​−sinθh​​ t=sinθ1​g2h​​

Q. A smooth inclined plane is inclined at an angle $θ$ with the horizontal. A body starts from rest and slides down the inclined surface, then the time taken by the body to reach the bottom is

$\frac{2 h}{g}$

$\frac{21}{g}$

$\frac{1}{s i n θ} \frac{2 h}{g}$

$sin θ \frac{2 h}{g}$