A body of mass 1mathrmkg$1\mathrm{kg}$ is suspended with the help of two strings making angles as shown in figure. Magnitude of tensions mathbfT_1$\mathbf{T}_1$ and mathbfT_2$\mathbf{T}_2$ , respectively, are (in N):
The diagram shows a suspended mass of 1 kg held by two strings making angles of 60 and 30 degrees with the horizontal.
A.5, 5sqrt3$5, 5\sqrt{3}$
B.5sqrt3, 5$5\sqrt{3}, 5$
C.5sqrt3, 5sqrt3$5\sqrt{3}, 5\sqrt{3}$
D.5, 5$5, 5$
Solution & Explanation
### Related Formula
For a system in static equilibrium:
sum F_x = 0 quad textand quad sum F_y = 0$$\sum F_x = 0 \quad \text{and} \quad \sum F_y = 0$$
### Core Logic
Let's resolve the tension forces vecT_1$\vec{T}_1$ and vecT_2$\vec{T}_2$ into horizontal and vertical components:
- T_1$T_1$ makes 60^circ$60^\circ$ with the horizontal.
- T_2$T_2$ makes 30^circ$30^\circ$ with the horizontal.
- Downward gravitational force: W = m g = 1 times 10 = 10 \ mathrmN$W = m g = 1 \times 10 = 10 \ \mathrm{N}$.
1. **Horizontal Equilibrium (sum F_x = 0$\sum F_x = 0$):**
T_1 cos 60^circ = T_2 cos 30^circ$$T_1 \cos 60^\circ = T_2 \cos 30^\circ$$T_1 cdot frac12 = T_2 cdot fracsqrt32 implies T_1 = T_2 sqrt3$$T_1 \cdot \frac{1}{2} = T_2 \cdot \frac{\sqrt{3}}{2} \implies T_1 = T_2 \sqrt{3}$$
2. **Vertical Equilibrium (sum F_y = 0$\sum F_y = 0$):**
T_1 sin 60^circ + T_2 sin 30^circ = m g = 10$$T_1 \sin 60^\circ + T_2 \sin 30^\circ = m g = 10$$T_1 cdot fracsqrt32 + T_2 cdot frac12 = 10$$T_1 \cdot \frac{\sqrt{3}}{2} + T_2 \cdot \frac{1}{2} = 10$$
### Step 1: Solve for Tensions
Substitute T_1 = T_2 sqrt3$T_1 = T_2 \sqrt{3}$ into the vertical equilibrium equation:
(T_2 sqrt3) fracsqrt32 + fracT_22 = 10$$(T_2 \sqrt{3}) \frac{\sqrt{3}}{2} + \frac{T_2}{2} = 10$$frac3 T_22 + fracT_22 = 10 implies 2 T_2 = 10 implies T_2 = 5 \ mathrmN$$\frac{3 T_2}{2} + \frac{T_2}{2} = 10 \implies 2 T_2 = 10 \implies T_2 = 5 \ \mathrm{N}$$
Substitute T_2$T_2$ back to obtain T_1$T_1$:
T_1 = 5 sqrt3 \ mathrmN$$T_1 = 5 \sqrt{3} \ \mathrm{N}$$
Thus, the tension magnitudes are T_1 = 5sqrt3 \ mathrmN$T_1 = 5\sqrt{3} \ \mathrm{N}$ and T_2 = 5 \ mathrmN$T_2 = 5 \ \mathrm{N}$.
### Pattern Recognition
Sees: Suspending particle static equilibrium with asymmetric strings.
Trap: Associating components with incorrect trigonometry axes or swapping T_1$T_1$ and T_2$T_2$ in options.
Shortcut: Since the incline of T_1$T_1$ (60^circ$60^\circ$) is steeper than that of T_2$T_2$ (30^circ$30^\circ$), T_1$T_1$ must carry a larger portion of the load, meaning T_1 > T_2$T_1 > T_2$. From the choices, only (2) satisfies this hierarchy.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
The diagram shows a suspended mass of 1 kg held by two strings making angles of 60 and 30 degrees with the horizontal.
A coin is placed on a disc. The coefficient of friction between the coin and the disc is mu$\mu$. If the distance of the coin from the center of the disc is r$r$, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is:
A.fracmu gr$\frac{\mu g}{r}$
B.sqrtfracrmu g$\sqrt{\frac{r}{\mu g}}$
C.sqrtfracmu gr$\sqrt{\frac{\mu g}{r}}$
D.fracmusqrtrg$\frac{\mu}{\sqrt{rg}}$
Solution
### Related Formula
f_s leq mu_s N$$f_s \leq \mu_s N$$F_c = mromega^2$$F_c = mr\omega^2$$
### Core Logic
Circular Motion Friction diagram for Q47 - JEE Main 2024 MorningCircular Motion Friction diagram for Q47 - JEE Main 2024 Morning
To prevent the coin from slipping, the static friction must provide the necessary centripetal force for circular motion.
f = momega^2 r$f = m\omega^2 r$
The normal force on the flat disc is N = mg$N = mg$.
The maximum static friction is f_textmax = mu N = mu mg$f_{\text{max}} = \mu N = \mu mg$.
For no slipping:
m r omega^2 leq mu mg$$m r \omega^2 \leq \mu mg$$omega^2 leq fracmu gr$$\omega^2 \leq \frac{\mu g}{r}$$omega_textmax = sqrtfracmu gr$$\omega_{\text{max}} = \sqrt{\frac{\mu g}{r}}$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws Of Motion
More Laws of Motion Questions — jee_main_2025_02_april_evening
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