Mechanical Properties of Solids and Fluids MCQ Questions & Answers in Basic Physics | Physics

$$\eqalign{ & Y = \left( {\frac{F}{A}} \right) \times \frac{L}{{\Delta L}} \cr & \Rightarrow \Delta L\left( {\frac{F}{A}} \right)\frac{L}{Y} = \frac{{8 \times {{10}^8} \times 0.5}}{{2 \times {{10}^{11}}}} = 2\,mm \cr} $$

As the weight $$w$$ balances the normal reactions.

So, $$w = {N_1} + {N_2}\,......\left( {\text{i}} \right)$$
Now balancing torque about the $$COM,$$
i.e. anti-clockwise momentum = clockwise momentum
$$ \Rightarrow {N_1}x = {N_2}\left( {d - x} \right)$$
Putting the value of $${N_2}$$ from Eq. (i), we get
$$\eqalign{ & {N_1}x = \left( {w - {N_1}} \right)\left( {d - x} \right) \cr & \Rightarrow {N_1}x = wd - wx - {N_1}d + {N_1}x \cr & \Rightarrow {N_1}d = w\left( {d - x} \right) \cr & \Rightarrow {N_1} = \frac{{w\left( {d - x} \right)}}{d} \cr} $$

As surface area decreases so energy is released.
Energy released $$ = 4\pi {R^2}T\left[ {{n^{\frac{1}{3}}} - 1} \right]\,\,\left( {{\text{where}}\,R = {n^{\frac{1}{3}}}r} \right)$$
$$\eqalign{ & = 4\pi {R^3}T\left[ {\frac{1}{r} - \frac{1}{R}} \right] \cr & = 3VT\left[ {\frac{1}{r} - \frac{1}{R}} \right] \cr} $$

According to Bernoulli’s theorem
$${P_1} + \frac{1}{2}\rho v_1^2 = {P_2} + \frac{1}{2}\rho v_2^2\,......\left( {\text{i}} \right)$$
From question,
$${P_1} - {P_2} = 3 \times {10^5},\frac{{{A_1}}}{{{A_2}}} = 5$$
According to equation of continuity
$${A_1}{v_1} = {A_2}{v_2}\,\,{\text{or,}}\,\,\frac{{{A_1}}}{{{A_2}}} = \frac{{{v_2}}}{{{v_1}}} = 5 \Rightarrow {v_2} = 5{v_1}$$
From equation (i)
$$\eqalign{ & {P_1} - {P_2} = \frac{1}{2}\rho \left( {v_2^2 - v_1^2} \right) \cr & {\text{or}}\,\,3 \times {10^5} = \frac{1}{2} \times 1000\left( {25v_1^2 - v_1^2} \right) \cr & \Rightarrow 600 = 6{v_1} \times 4{v_1} \Rightarrow v_1^2 = 25 \cr & \therefore {v_1} = 5\,m/s \cr} $$

From the graph $$l = {10^{ - 4}}m,F = 20\,N$$
$$\eqalign{ & A = {10^{ - 6}}{m^2},L = 1m \cr & \therefore Y = \frac{{FL}}{{Al}} = \frac{{20 \times 1}}{{{{10}^{ - 6}} \times {{10}^{ - 4}}}} \cr & = 20 \times {10^{10}} \cr & = 2 \times {10^{11}}\,N/{m^2} \cr} $$

$${\text{strain}} = \frac{F}{{\pi {r^2}\gamma }} = {10^{ - 8}}$$

Let $${a_1}$$ and $${a_2}$$ be cross sectional areas and $${v_1}$$ and $${v_2}$$ the velocities of liquid flow at $$A$$ and $$B.$$ If $${p_1}$$ and $${p_2}$$ are pressures of liquid recorded by manometer, then
$$\eqalign{ & {p_1} + \frac{1}{2}pv_1^2 = {p_2} + \frac{1}{2}\rho v_2^2 \cr & \Rightarrow {p_1} - {p_2} = \frac{1}{2}\rho v_1^2\left( {\frac{{v_2^2}}{{v_1^2}} - 1} \right) \cr} $$
By equation of continuity, $${a_1}{v_1} = {a_2}{v_2}$$
Also, $${p_1} - {p_2} = h\rho g$$
Substituting these values, we have
$${v_1} = \sqrt {\frac{{2gh}}{{\frac{{a_1^2}}{{a_2^2}} - 1}}} $$
Rate of flow of liquid is $${a_1}{v_1} = {a_2}{a_2}\sqrt {\frac{{2gh}}{{a_1^2 - a_2^2}}} $$

$$\eqalign{ & {\text{Ratio of radii }}{r_1}:{r_2} = 1:2 \cr & {\text{Ratio of area,}}\,{A_1}:{A_2} = \pi r_1^2:\pi r_2^2 \cr & {A_1}:{A_2} = 1:4 \cr & {\text{Now, Stres}}{{\text{s}}_1}:{\text{Stres}}{{\text{s}}_2} = {\text{ }}4:1 \cr & {\text{So,}}\,{\text{Strai}}{{\text{n}}_1}:{\text{Strai}}{{\text{n}}_2} = 4:1 \cr & \therefore \frac{{{l_1}}}{{{l_2}}} = \frac{4}{1} \Rightarrow 4{l_2} = {l_1} = 8 \cr & \therefore {l_2} = 2\,mm \cr & {\text{Increase in length of }}B{\text{ is }}2\,mm. \cr} $$

From the equation of continuity,
$$\eqalign{ & {\text{Rate}} = r = \frac{A}{3}{V_p} + \left( {A - \frac{A}{3}} \right){V_Q} \cr & \therefore {V_P} + 2\;{V_Q} = \frac{{3r}}{A} = \frac{{3 \times 0.1}}{{{{10}^{ - 2}}}} = 30\,m/s \cr & {\text{As}}\,{V_P} = 20\,m/s,\,{\text{so}}\,{V_Q} = 5\,m/s \cr} $$

Couple per unit angle of twist, $$C = \frac{{\pi \eta {r^4}}}{{2\ell }}$$
$$\therefore $$ Couple $$\tau = C\theta = \frac{{\pi \eta {r^4}\theta }}{{2\ell }}$$
Here $$\eta ,\ell ,C\,\& \,\tau $$   are same. So, $${r^4}\theta = {\text{constant}}$$
$$\therefore \frac{{{\theta _1}}}{{{\theta _2}}} = \left( {\frac{{r_2^4}}{{r_1^4}}} \right)$$