A body of mass 1kg is suspended with the help of two strings making angles as shown in the figure. Magnitude of tensions $ T_1 $ and $ T_2 $, respectively, are (in N):
A body of mass 1kg is suspended with the help of two strings making angles as shown in the figure. Magnitude of tensions $ T_1 $ and $ T_2 $, respectively, are (in N):
Show Hint
- 5, \( 5\sqrt{3} \)
- \( 5\sqrt{3} \), 5
- \( 5\sqrt{3} \), \( 5\sqrt{3} \)
- 5, 5
The Correct Option is B
Approach Solution - 1
Given that the body is in equilibrium, we can resolve the forces in the vertical and horizontal directions.
The weight of the body is \( mg = 1 \times 9.8 = 9.8 \, \text{N} \). For the vertical direction: \[ T_1 \sin 30^\circ + T_2 \sin 30^\circ = mg \] For the horizontal direction: \[ T_1 \cos 30^\circ = T_2 \cos 30^\circ \] Thus: \[ T_1 = T_2 \] Now, solving for the tensions using the vertical direction equation: \[ T_1 \sin 30^\circ + T_1 \sin 30^\circ = 9.8 \, \text{N} \] \[ 2T_1 \sin 30^\circ = 9.8 \] \[ 2T_1 \times \frac{1}{2} = 9.8 \] \[ T_1 = 5 \, \text{N}, \, T_2 = 5\sqrt{3} \, \text{N} \] Thus, the correct answer is: \( T_1 = 5 \), \(\text{N} \, T_2 = 5\sqrt{3} \), \(\text{N} \)
Approach Solution -2
A 1 kg mass is suspended by two strings making angles of \(60^\circ\) (left) and \(30^\circ\) (right) with the horizontal. Let the tensions be \(T_1\) (left string) and \(T_2\) (right string). We need the magnitudes of \(T_1\) and \(T_2\).
Concept Used:
For a body in equilibrium, the vector sum of forces is zero. Resolve each tension into horizontal and vertical components and apply:
\[ \sum F_x=0,\qquad \sum F_y=0. \]Take \(g=10~\text{m s}^{-2}\Rightarrow mg=10~\text{N}\) (standard JEE convention).
Step-by-Step Solution:
Step 1: Write components of tensions (angles with the horizontal):
\[ T_1:\ \begin{cases} \text{horizontal (left)} = T_1\cos 60^\circ,\\ \text{vertical (up)} = T_1\sin 60^\circ, \end{cases}\qquad T_2:\ \begin{cases} \text{horizontal (right)} = T_2\cos 30^\circ,\\ \text{vertical (up)} = T_2\sin 30^\circ. \end{cases} \]Step 2: Horizontal equilibrium (\(\sum F_x=0\)):
\[ T_2\cos 30^\circ = T_1\cos 60^\circ \;\Rightarrow\; T_2\left(\frac{\sqrt{3}}{2}\right)=T_1\left(\frac{1}{2}\right) \;\Rightarrow\; T_1=\sqrt{3}\,T_2. \]Step 3: Vertical equilibrium (\(\sum F_y=0\)):
\[ T_1\sin 60^\circ + T_2\sin 30^\circ = mg \;\Rightarrow\; T_1\left(\frac{\sqrt{3}}{2}\right) + T_2\left(\frac{1}{2}\right) = 10. \] \[ \text{Using } T_1=\sqrt{3}\,T_2:\quad \frac{\sqrt{3}}{2}(\sqrt{3}T_2)+\frac{1}{2}T_2 = \left(\frac{3}{2}+\frac{1}{2}\right)T_2 = 2T_2 = 10 \;\Rightarrow\; T_2=5~\text{N}. \]Step 4: Find \(T_1\):
\[ T_1=\sqrt{3}\,T_2=\sqrt{3}\times 5=5\sqrt{3}~\text{N}. \]Final Computation & Result:
The magnitudes of tensions are
\[ \boxed{T_1=5\sqrt{3}\ \text{N},\quad T_2=5\ \text{N}} \]Hence, the required pair is \( (5\sqrt{3},\,5) \) N.
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