A body of weight $2 ext{ kg}$ is suspended as shown in the figure. The tension $T_1$ in the horizontal strin

Question

A body of weight $2 	ext{ kg}$ is suspended as shown in the figure. The tension $T_1$ in the horizontal string (in $	ext{kg wt}$) is:

Accepted Answer

1. **Equilibrium Principle:** For the knot connecting the strings to remain in equilibrium, the vector sum of all forces acting on it must be zero [NCERT Class 11, Physics Part I, Section 5.8].
2. **Free Body Diagram Analysis:**
   - **Downward force:** Weight $W = 2 	ext{ kg wt}$.
   - **Horizontal force:** Tension $T_1$.
   - **Inclined force:** Tension $T_2$ in the string attached to the support.
3. **Resolution of Forces:** Let the inclined string make an angle $	heta$ with the vertical. Resolving $T_2$ into rectangular components:
   - The vertical component balances the weight: $T_2 \cos 	heta = W = 2$.
   - The horizontal component balances the tension $T_1$: $T_2 \sin 	heta = T_1$.
4. **Calculation:** Dividing the horizontal equation by the vertical equation eliminates $T_2$:
   $$ \frac{T_1}{2} = \frac{T_2 \sin 	heta}{T_2 \cos 	heta} = 	an 	heta $$
   $$ T_1 = 2 	an 	heta $$
5. **Deduction:** The probable answer $2\sqrt{3}$ implies that $	an 	heta = \sqrt{3}$, which corresponds to $	heta = 60^\circ$ (angle with the vertical). Assuming this standard configuration from the missing figure, the calculated tension is $2\sqrt{3} 	ext{ kg wt}$.

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