A heavy iron bar, of weight W is having its one end on the ground and the other on the shoulder of a person. The bar makes an angle \(\theta\) with the horizontal. The weight experienced by the person is:
- \(\frac{W}{2}\)
- W
- \(W cos\theta\)
- \(W sin\theta\)
The Correct Option is A
Approach Solution - 1
The problem at hand involves analyzing the distribution of weight when a heavy iron bar is balanced such that one end is on the ground and the other is resting on a person's shoulder. Here is how we can solve this:
Understanding the Situation:
- The iron bar is static, balanced between the ground and the person's shoulder, making an angle \(\theta\) with the horizontal.
- This is a standard physics problem involving equilibrium and the distribution of forces.
Principle Involved:
- When a uniform rod is balanced at two points, the weight of the rod is distributed equally at those two points when the rod is symmetrically placed.
- Therefore, the total weight \(W\) is shared equally between the contact points, which are the ground and the shoulder of the person.
Solution Steps:
- Consider the iron bar as a uniform rod. The gravitational force acts at its center of mass, which is at the midpoint of the rod.
- Since the rod is symmetric and in equilibrium, the supported forces by the ground and the person must be equal to balance the gravitational force acting on it.
- This results in each support (ground and shoulder) experiencing half of the total weight of the bar.
- Hence, the weight experienced by the person is \(\frac{W}{2}\).
Conclusion:
- The correct answer is: \(\frac{W}{2}\) as the weight of the rod is equally distributed between the ground and the person.
Thus, the weight experienced by the person is option: \(\frac{W}{2}\).
Approach Solution -2
Step 1: Concept of center of mass The weight of the bar is uniformly distributed. Since the bar is uniform, the center of mass lies at the midpoint of the bar.
The total weight \( W \) of the bar is supported by two points:
- One end is on the ground.
- The other end is on the shoulder of the person.
Step 2: Distribution of weight In a symmetric situation like this, the total weight is evenly distributed between the two points of contact.
\[ \text{Weight experienced by the person} = \frac{W}{2}. \]
Final Answer: \( \frac{W}{2} \).
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Forces Questions
- A block of mass M slides down on a rough inclined plane with constant velocity. The angle made by the inclined plane with horizontal is θ.The magnitude of the contact force will be
A body of mass of \(4\;kg\) experiences two forces \(\vec{F_1}=5\hat i+8\hat j+7\hat k \) and \(\vec{F_2}=3\hat i-4\hat j-3\hat k\) then acceleration acting on the body \(R\)
- A uniform wire has length L and radius r. It is acted on by a force F as shown. The elongation is l. If F and r are both halved, the new elongation will be :
- Two forces \(F_1\) and \(F_2\) are applied on two rods \(P\) and \(Q\) of same materials such that elongation in rods are same. If ratio of their radii is \(x : y\) and ratio of length is \(m : n\), then ratio of \(F_1 : F_2\) is
- For the block shown, \(F_1\) is the minimum force required to move block upward and \(F_2\) is the minimum force required to prevent it from slipping find \(| F_1 - F_2|\)
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.