The coordinates of a particle with respect to origin in a given reference frame is \( (1, 1, 1) \) meters. If a force of \( \mathbf{F} = \hat{i} - \hat{j} + \hat{k} \) acts on the particle, then the magnitude of torque (with respect to origin) in the z-direction is:
Correct Answer: 2
Approach Solution - 1
To find the torque \( \mathbf{\tau} \) exerted by the force \( \mathbf{F} = \hat{i} - \hat{j} + \hat{k} \) on the particle at location \( \mathbf{r} = (1 \,\text{m}, 1 \,\text{m}, 1 \,\text{m}) \) with respect to the origin, we use the cross product formula for torque: \( \mathbf{\tau} = \mathbf{r} \times \mathbf{F} \).
We can express \( \mathbf{r} \) and \( \mathbf{F} \) as vectors:
\(\mathbf{r} = \hat{i} + \hat{j} + \hat{k}, \quad \mathbf{F} = \hat{i} - \hat{j} + \hat{k}\)
The cross product \(\mathbf{r} \times \mathbf{F}\) is computed using the determinant formula:
\(\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\1 & 1 & 1 \\1 & -1 & 1\end{vmatrix}\)
The determinant can be expanded as follows:
- \(\hat{i}(1 \cdot 1 - 1 \cdot (-1)) = \hat{i}(1 + 1) = 2\hat{i}\)
- \(\hat{j}(-(1 \cdot 1 - 1 \cdot 1)) = \hat{j}(0) = 0\hat{j}\)
- \(\hat{k}(1 \cdot (-1) - 1 \cdot 1) = \hat{k}(-1 - 1) = -2\hat{k}\)
Therefore, the torque vector is:
\(\mathbf{\tau} = 2\hat{i} + 0\hat{j} - 2\hat{k}\)
The magnitude of the torque in the z-direction is represented by the coefficient of \(\hat{k}\), which is \( |-2| = 2 \).
Approach Solution -2
Position vector of the particle:
\[ \vec{r} = (1\hat{i} + 1\hat{j} + 1\hat{k}) \, \text{m} \]
Force acting on the particle:
\[ \vec{F} = (\hat{i} - \hat{j} + \hat{k}) \, \text{N} \]
Step 2: Formula for torque.
The torque of a force about the origin is given by the vector cross product:
\[ \vec{\tau} = \vec{r} \times \vec{F} \] We can compute this using the determinant form:
\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix} \]
Step 3: Expand the determinant.
\[ \vec{\tau} = \hat{i}[(1)(1) - (1)(-1)] - \hat{j}[(1)(1) - (1)(1)] + \hat{k}[(1)(-1) - (1)(1)] \] Simplify each term:
\[ \vec{\tau} = \hat{i}(2) - \hat{j}(0) + \hat{k}(-2) \] \[ \vec{\tau} = 2\hat{i} - 2\hat{k} \]
Step 4: Find the component in the z-direction.
The z-component of torque is the coefficient of \( \hat{k} \):
\[ \tau_z = -2 \] Therefore, the magnitude of the torque in the z-direction is:
\[ |\tau_z| = 2 \]
Final Answer:
\[ \boxed{2} \]
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 torque Questions
- A uniform magnetic field of \( 2 \times 10^{-3} \, \text{T} \) acts along positive Y-direction. A rectangular loop of sides 20 cm and 10 cm with current of 5 A is in the Y-Z plane. The current is in anticlockwise sense with reference to the negative X-axis. Magnitude and direction of the torque is:
- A uniform rod of mass 250 g having length 100 cm is balanced on a sharp edge at the 40 cm mark. A mass of 400 g is suspended at the 10 cm mark. To maintain the balance of the rod, the mass to be suspended at the 90 cm mark is:
- The coordinates of a particle with respect to origin in a given reference frame is \( (1, 1, 1) \) meters. If a force of \( \mathbf{F} = \hat{i} - \hat{j} + \hat{k} \) acts on the particle, then the magnitude of torque (with respect to origin) in the z-direction is:
- A cube having a side of 10 cm with unknown mass and 200 gm mass were hung at two ends of an uniform rigid rod of 27 cm long. The rod along with masses was placed on a wedge keeping the distance between wedge point and 200 gm weight as 25 cm. Initially the masses were not at balance. A beaker is placed beneath the unknown mass and water is added slowly to it. At given point the masses were in balance and half volume of the unknown mass was inside the water. (Take the density of the unknown mass is more than that of the water, the mass did not absorb water and water density is 1 gm/cm$^3$.) The unknown mass is ______ kg.
A square Lamina OABC of length 10 cm is pivoted at \( O \). Forces act at Lamina as shown in figure. If Lamina remains stationary, then the magnitude of \( F \) is:
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 ______.