Two metal spheres, one of radius R and the other of radius 2R respectively have the same surface charge density σ. They are brought in contact and separated. What will be the new surface charge densities on them ?

Suppose the charge of a proton and an electron differ slightly. One of them is - e, the other is (e + ∆e). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then ∆e is of the order of [Given mass of hydrogen mh = 1.67 × 10−27 kg ] ?

If the work done is stretching a wire by 1 mm is 2 J, the work necessary for stretching another wire of same material but with double the radius of cross section and half the length by 1 mm is

Two identical charged spheres suspended from a common point by two massless strings of lengths l, are initially at a distance d(d ≪ l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres, as: ?

Two pith balls carrying equal charges are suspended from a common point by strings of equal length. The equilibrium separation between them is r. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now become ?

A charge ’ q ’ is placed at the centre of the line joining two equal charges ’ Q ’. The system of the three charges will be in equilibrium if ’ q ’ is equal to ?

Two metallic spheres of radii 1 cm and 3 cm are given charges of −1 × $10^{−2}$C and 5 × $10^{−2}$C, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is ?

A 3 kg ball strikes a heavy rigid wall with a speed of 10 m/s at an angle of 60◦. It gets reflected with the same speed and angle as shown here. If the ball is in contact with the wall for 0.20 s, what is the average force exerted on the ball by the wall?

A satellite in a force free space sweeps stationary interplanetary dust at a rate (dM/dt) = αv. The acceleration of satellite is

Physical independence of force is a consequence of

A particle of mass m is moving with a uniform velocity $v_1$. It is given an impulse such that its velocity becomes $v_2$. The impulse is equal to

A 600 kg rocket is set for a vertical firing. If the exhaust speed is 1000 $ms^{−1}$, the mass of the gas ejected per second to supply the thrust needed to overcome the weight of rocket is

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\[ \frac{F_{\text{electric}}}{F_{\text{gravity}}}\]



\[ \frac{F_{\text{electric}}}{F_{\text{gravity}}} = \frac{\frac{1}{4\pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2}}{G \cdot \frac{m_1 m_2}{r^2}} = \frac{(9 \times 10^9) \cdot \frac{(1.6 \times 10^{-19})(1.6 \times 10^{-19})}{r^2}}{(6.67 \times 10^{-11}) \cdot \frac{(1.67 \times 10^{-27})(9.11 \times 10^{-31})}{r^2}} = 2.4 \times 10^{39} \] \[ I = m_1 x^2 + m_2 (L - x)^2 \] \[ \frac{dI}{dx} = m_1 \cdot 2x + m_2 \cdot 2(L - x)(0 - 1) \] \[ \text{Minimum when } \frac{dI}{dx} = 0 \] \[ m_1 \cdot 2x = m_2 \cdot 2(L - x) \] \[ m_1 \cdot 2x = m_2 \cdot 2L - m_2 \cdot 2x \] \[ (m_1 + m_2) \cdot 2x = m_2 \cdot 2L \] \[ \Rightarrow x = \frac{m_2 L}{m_1 + m_2} \] 

\[ \frac{\Delta L}{L} = ? \] \[ \text{Given: } \text{Hypotenuse } = L + \Delta L, \quad \text{Opposite side } = x \] Using Pythagoras’ Theorem: \[ L^2 + x^2 = (L + \Delta L)^2 \] \[ L^2 + x^2 = \left( L \left(1 + \frac{\Delta L}{L} \right) \right)^2 \] \[ L^2 + x^2 = L^2 \left(1 + \frac{\Delta L}{L} \right)^2 \] Divide both sides by \( L^2 \): \[ 1 + \frac{x^2}{L^2} = \left(1 + \frac{\Delta L}{L} \right)^2 \] \[ \sqrt{1 + \frac{x^2}{L^2}} = 1 + \frac{\Delta L}{L} \] Using the approximation \( \sqrt{1 + x} \approx 1 + \frac{x}{2} \) for small \( x \), we get: \[ 1 + \frac{x^2}{2L^2} \approx 1 + \frac{\Delta L}{L} \] \[ \Rightarrow \frac{\Delta L}{L} \approx \frac{x^2}{2L^2} \] \[ \boxed{\frac{\Delta L}{L} = \frac{x^2}{2L^2}} \]



\[ \sum F = ma \]\[ 100 - \mu mg = ma \] \[ 100 - (0.5)(10)(10) = (10)a \] \[ 50 = 10\hspace{2mm}a \] \[ a = 5 \, \text{m/s}^2 \]