A force\(- F \hat k\) acts on O, the origin of the coordinate system. The torque at the point (1, -1) will be:

1. $-\mathrm{F}\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}\right)$

2. $\mathrm{F}\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}\right)$

3. $-\mathrm{F}\left(\hat{\mathrm{i}}-\hat{\mathrm{j}}\right)$

4. $\mathrm{F}\left(\hat{\mathrm{i}}-\hat{\mathrm{j}}\right)$

A thin uniform circular disc of mass \(M\) and radius \(R\) is rotating in a horizontal plane about an axis passing through its center and perpendicular to its plane with an angular velocity $\omega$. Another disc of the same dimensions but of mass \(\frac{1}{4}M\) is placed gently on the first disc co-axially. The angular velocity of the system will be:

A particle of mass \(m\) moves in the\(XY\) plane with a velocity of \(v\) along the straight line \(AB.\) If the angular momentum of the particle about the origin \(O\) is \(L_A\) when it is at \(A\) and \(L_B\) when it is at \(B,\) then:

A wheel with a radius of \(20\) cm has forces applied to it as shown in the figure. The torque produced by the forces of \(4\) N at \(A\), \(8~\)N at \(B\), \(6\) N at \(C\), and \(9~\)N at \(D\), at the angles indicated, is:

1. \(5.4\) N-m anticlockwise

2. \(1.80\) N-m clockwise

3. \(2.0\) N-m clockwise

4. \(3.6\) N-m clockwise

If the radius of the earth is suddenly contracted to half of its present value, then the duration of the day will be of:

The centre of the mass of \(3\) particles, \(10\) kg, \(20\) kg, and \(30\) kg, is at \((0,0,0)\). Where should a particle with a mass of \(40\) kg be placed so that its combined centre of mass is \((3,3,3)\)?
1. \((0,0,0)\)
2. \((7.5, 7.5, 7.5)\)
3. \((1,2,3)\)
4. \((4,4,4)\)

The position of a particle is given by \(\vec r = \hat i+2\hat j-\hat k\) and momentum \(\vec P = (3 \hat i + 4\hat j - 2\hat k)\). The angular momentum is perpendicular to:

A rigid body rotates about a fixed axis with a variable angular velocity equal to \(\alpha\) \(-\) \(\beta t\), at the time \(t\), where \(\alpha , \beta\) are constants. The angle through which it rotates before it stops is:

Four particles of mass \(m_1 = 2m\), \(m_2=4m\), \(m_3 =m \), and \(m_4\) are placed at the four corners of a square. What should be the value of \(m_4\) so that the centre of mass of all the four particles is exactly at the centre of the square?