Q151 / 180 · COMEDK 2026 (9 May Shift 2) · Physics
A block of metal, of 25g mass moves down without acceleration when the plane is inclined at an angle of $30^\circ$. When the inclination is increased by $30^\circ$, find the downward acceleration of the block.
- Correct answerA$5.66\ ms^{-2}$
- B$3.8\ ms^{-2}$
- C$2.6\ ms^{-2}$
- D$1.9\ ms^{-2}$
Solution
When the block moves down without acceleration at an angle $\theta = 30^\circ$, the net force is zero.
$mg \sin 30^\circ = \mu_k mg \cos 30^\circ$
$\mu_k = \tan 30^\circ = \dfrac{1}{\sqrt{3}}$
When the inclination is increased by $30^\circ$, the new angle is $\theta' = 60^\circ$.
The downward acceleration $a$ is given by:
$ma = mg \sin 60^\circ - \mu_k mg \cos 60^\circ$
$a = g \left( \sin 60^\circ - \mu_k \cos 60^\circ \right)$
Substituting the values of $\mu_k$ and $g = 9.8\ ms^{-2}$:
$a = 9.8 \left( \dfrac{\sqrt{3}}{2} - \dfrac{1}{\sqrt{3}} \left(\dfrac{1}{2}\right) \right)$
$a = 9.8 \left( \dfrac{3 - 1}{2\sqrt{3}} \right) = \dfrac{9.8}{\sqrt{3}}$
$a = \dfrac{9.8}{1.732} \approx 5.66\ ms^{-2}$
Answer: $5.66\ ms^{-2}$
$mg \sin 30^\circ = \mu_k mg \cos 30^\circ$
$\mu_k = \tan 30^\circ = \dfrac{1}{\sqrt{3}}$
When the inclination is increased by $30^\circ$, the new angle is $\theta' = 60^\circ$.
The downward acceleration $a$ is given by:
$ma = mg \sin 60^\circ - \mu_k mg \cos 60^\circ$
$a = g \left( \sin 60^\circ - \mu_k \cos 60^\circ \right)$
Substituting the values of $\mu_k$ and $g = 9.8\ ms^{-2}$:
$a = 9.8 \left( \dfrac{\sqrt{3}}{2} - \dfrac{1}{\sqrt{3}} \left(\dfrac{1}{2}\right) \right)$
$a = 9.8 \left( \dfrac{3 - 1}{2\sqrt{3}} \right) = \dfrac{9.8}{\sqrt{3}}$
$a = \dfrac{9.8}{1.732} \approx 5.66\ ms^{-2}$
Answer: $5.66\ ms^{-2}$
Hence, the correct answer is option: 'A'.