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NVAEasyJEE 2023Newton's Second Law & Force

JEE Physics 2023 Question with Solution

Banner image for JEE Main 2023 question paper with NTA logo and session details, not part of the physics problem.

Physics Section-A

A block of 3 kg\sqrt{3} \, \text{kg} is attached to a string whose other end is attached to the wall. An unknown force FF is applied so that the string makes an angle of 30∘30^\circ with the wall. The tension TT is:

(Given g=10 ms−2g = 10 \, \text{ms}^{-2})

A wall on the left supports a string making 30 degree angle with the wall, joined to a junction pulled right by force F and downward by a block of mass root 3 kg.

Answer

Correct answer:20

Step-by-step solution

Standard Method

Given: A block of mass 3 kg\sqrt{3} \, \text{kg} is hanging from the junction, the string makes an angle θ=30∘\theta = 30^\circ with the wall, and g=10 ms−2g = 10 \, \text{ms}^{-2}.

Find: The tension TT in the inclined string.

Free body style diagram showing inclined tension T at 30 degree to the wall, horizontal force F to the right, and downward load equal to root 3 g.

From the solution working,

cos⁡θ=3gT\cos \theta = \frac{\sqrt{3}g}{T}

With θ=30∘\theta = 30^\circ,

32=3gT\frac{\sqrt{3}}{2} = \frac{\sqrt{3}g}{T}

Using g=10 ms−2g = 10 \, \text{ms}^{-2},

32=103T\frac{\sqrt{3}}{2} = \frac{10\sqrt{3}}{T}

So,

T=20 NT = 20 \, \text{N}

Therefore, the tension is 20 N20 \, \text{N}.

Component Balance

Given: The junction is in equilibrium under three forces: the tension TT along the string, the horizontal applied force FF, and the downward pull due to the hanging block.

Find: The value of TT.

The weight of the block is

W=mg=3×10=103 NW = mg = \sqrt{3} \times 10 = 10\sqrt{3} \, \text{N}

Since the string makes 30∘30^\circ with the vertical wall, its vertical component is

Tcos⁡30∘T \cos 30^\circ

For vertical equilibrium,

Tcos⁡30∘=103T \cos 30^\circ = 10\sqrt{3}

Now substitute cos⁡30∘=32\cos 30^\circ = \frac{\sqrt{3}}{2}:

T⋅32=103T \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3}

Cancelling 3\sqrt{3} from both sides,

T2=10\frac{T}{2} = 10

Hence,

T=20 NT = 20 \, \text{N}

Therefore, the required numerical value is 20.

Common mistakes

  • Using Tsin⁡30∘T \sin 30^\circ for the vertical component is incorrect because the angle is given with the wall, that is, with the vertical. Use Tcos⁡30∘T \cos 30^\circ for the vertical balance instead.

  • Taking the mass 3 kg\sqrt{3} \, \text{kg} directly as the downward force is incorrect. First convert mass to weight using W=mgW = mg, so the downward force is 103 N10\sqrt{3} \, \text{N}.

  • Writing the final answer as 20 N20 \, \text{N} in the answer field would be wrong for a numerical value answer. The stored answer should be only the number 20 without units.

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