Question

A particle slides down from rest on an inclined plane of angle $\theta$ with horizontal. The distances are as shown. The particle slides down from top to position A where its velocity is v. Match the options of two columns.

$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{(a)}& (v^2-2gh)\text{will}& \text{(p)}& \text{Increase}\\ \text{(b)}& (v^2-2gssin\theta )\text{will}& \text{(q)}& \text{Decrease}\\ \text{(c)}& (v^2-2g\frac{H}{p})\text{will}& \text{(r)}& \text{Remains constant}\\ \text{(d)}& [v^2-\frac{2gs(H-h)}{(p-s)}]\text{will}& \text{(s)}& \text{May increase or decrease}\end{array}$

• A. A-r; B-r; C-r; D-r
• B. A-r; B-s; C-q; D-p
• C. A-p; B-q; C-r; D-s
• D. A-q; B-r; C-r; D-q

Answer 1

The correct option is A

A-r; B-r; C-r; D-r

$h=ssin\theta ;\frac{1}{2}m{v}^2+mv(H-h)$ = constant

or $\frac{1}{2}m{v}^2-mgh=$ constant, as mgH is constant.

$(v^2–2gh)=$ constant; $v^2-2gssin\theta$ = constant.

But $sin\theta =\frac{H}{p}=\frac{(H-h)}{(p-s)}$