A slider block A moves downward at a speed of v_A = 2 m/s , at an… - Vidyadip

MCQ

A slider block $A$ moves downward at a speed of $v_A = 2\ m/s$ , at an angle of $75^o$ with horizontal as shown in the figure. The velocity with respect to $A$ of the portion of belt $B$ between ideal pulleys $C$ and $D$ is $v_{CD/A} = 2\ m/s$ at an angle $\theta $ with the horizontal. The magnitude of velocity of portion $CD$ of the belt when $\theta = 15^o$ is .......... $m/s$

A
$2\sqrt 3$
B
$\sqrt {10} $
C
$2\sqrt 2$
✓
$2$

✓

Answer

Correct option: D.

$2$

d
$\overrightarrow{\mathrm{V}}_{\mathrm{CD} / \mathrm{A}}=\overrightarrow{\mathrm{V}}_{\mathrm{CD}}-\overrightarrow{\mathrm{V}}_{\mathrm{A}}$

$\overrightarrow{\mathrm{V}}_{\mathrm{CD}}=\overrightarrow{\mathrm{V}}_{\mathrm{CD} / \mathrm{A}}+\overrightarrow{\mathrm{V}}_{\mathrm{A}}$

$\mathrm{V}_{\mathrm{CD}}^{2}=\mathrm{V}_{\mathrm{A}}^{2}+\mathrm{V}_{\mathrm{CD} / \mathrm{A}}^{2}-2 \mathrm{V}_{\mathrm{A}} \mathrm{V}_{\mathrm{CD} / \mathrm{A}} \cos 60^{\circ}$

$\mathrm{V}_{\mathrm{CD}}=2 \mathrm{m} / \mathrm{s}$

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