A box of mass 5 kg is pulled by a cord, up along a frictionless plane inclined at $30^{\circ}$ with the horizontal. The tension in the cord is 30 N . The acceleration of the box is (Take $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ )

• A

  $2 \mathrm{~m} \mathrm{~s}^{-2}$
• B

  Zero
• C

  $0.1 \mathrm{~m} \mathrm{~s}^{-2}$
• Correct answer
  D

  $1 \mathrm{~m} \mathrm{~s}^{-2}$

Solution

$T-M g \sin 30=M a$
$30-5 \times 10 \times \frac{1}{2}=5 a$ $a=1 \mathrm{~m} / \mathrm{s}^2$

Hence, the correct answer is option: 'D'.