A ball is placed on a wedge as shown in the figure. The wedge starts moving to the left with acceleration 2 {\g} \tan \\theta. The time taken by the ball to cover a distance {\L} along the plane is

Correct Answer is : t=√2Lgsinθ

t = \frac{2\L}{\g \sin \\theta}

t = \frac{\sqrt2\L}{\g \sin \\theta}

\t = \sqrt{{2\L} / {\g \sin \\theta}}

t = \frac{\g \sin \\theta}{2\L}

TS EAMCET Medical 30 Jul 2022 Shift 2 Paper

Question : 87 of 160

Marks: +1, -0

A ball is placed on a wedge as shown in the figure. The wedge starts moving to the left with acceleration 2g‌tan‌θ. The time taken by the ball to cover a distance L along the plane is

t =
2L
g‌sin‌θ
t =
√2L
g‌sin‌θ
t=√
2L
g‌sin‌θ
t =
g‌sin‌θ
2L

Validate

Go to Question:

Prev Question

Next Question