2.
If the ratio of radii of two wires of same material is $$3 : 1$$ and ratio of their lengths is $$5 : 1,$$ then the ratio of the normal forces that will produce the same extension in the length of two wires is
3.
A glass tube of uniform internal radius ($$r$$) has a valve separating the two identical ends. Initially, the valve is in a tightly closed position.
End $$1$$ has a hemispherical soap bubble of radius $$r.$$ End $$2$$ has sub-hemispherical soap bubble as shown in figure. Just after opening the valve,
A.
air from end $$1$$ flows towards end $$2.$$ No change in the volume of the soap bubbles
B.
air from end $$1$$ flows towards end $$2.$$ Volume of the soap bubbles at end $$1$$ decreases
C.
no changes occurs
D.
air from end $$2$$ flows towards end $$1.$$ Volume of the soap bubble at end $$1$$ increases
Answer :
air from end $$1$$ flows towards end $$2.$$ Volume of the soap bubbles at end $$1$$ decreases
We know that excess pressure in a soap bubble is inversely proportional to its radius. The soap bubble at end 1 has small radius as compared to the soap bubble at end 2 (given). Therefore excess pressure at 1 is more.
As the value is opened, air flows from end 1 to end 2 and the volume of soap bubble at end 1 decreases.
4.
A body $$B$$ is capable of remaining stationary inside a liquid at the position shown in Fig. (a). If the whole system is gently placed on smooth inclined plane (Fig (b)) and is allowed to slide down, then $$\left( {0 < \theta < {{90}^ \circ }} \right).$$ The body will
A.
move up (relative to liquid)
B.
move down (relative to liquid)
C.
remain stationary (relative to liquid)
D.
move up for some inclination $$\theta $$ and will move down for another inclination $$\theta $$
Answer :
move up for some inclination $$\theta $$ and will move down for another inclination $$\theta $$
6.
A platform is suspended by four wires at its corners. The wires are $$3m$$ long and have a diameter of $$2.0\,mm.$$ Young’s modulus for the material of the wires is $$1,80,000\,MPa.$$ How far will the platform drop (due to elongation of the wires) if a $$50\,kg$$ load is placed at the centre of the platform?
$$\eqalign{
& \Delta L = \frac{{LF}}{{AY}},\,{\text{where}}\,L = 3\,m, \cr
& A = \pi {\left( {1.0 \times {{10}^{ - 3}}\,m} \right)^2} = 3.14 \times {10^{ - 6}}\,{m^2}, \cr} $$
and since each wire supports one-quarter of the load,
$$\eqalign{
& F = \frac{{\left( {50\;kg} \right)\left( {9.8\,m/{s^2}} \right)}}{4} = 123\,N \cr
& \Delta L = \frac{{\left( {3\;m} \right)\left( {123\,N} \right)}}{{\left( {3.14 \times {{10}^{ - 6}}\,{m^2}} \right)\left( {1.8 \times {{10}^{11}}\,N/{m^2}} \right)}} \cr
& = 65 \times {10^{ - 5}}\,m\,\,{\text{or}}\,\,0.65\,mm \cr} $$
7.
When a force is applied on a wire of uniform cross-section area $$3 \times {10^{ - 6}}{m^2}$$ and length $$4m,$$ the increase in length is $$1\,mm.$$ Energy stored in it will be $$\left( {Y = 2 \times {{10}^{11}}\,N/{m^2}} \right)$$
8.
A large number of droplets, each of radius, $$r$$ coalesce to form a bigger drop of radius, $$R.$$ An engineer designs a machine so that the energy released in this process is converted into the kinetic energy of the drop. Velocity of the drop is ($$T$$ = surface tension, $$\rho $$ = density)
A.
$${\left[ {\frac{T}{\rho }\left( {\frac{1}{r} - \frac{1}{R}} \right)} \right]^{\frac{1}{2}}}$$
B.
$${\left[ {\frac{{6T}}{\rho }\left( {\frac{1}{r} - \frac{1}{R}} \right)} \right]^{\frac{1}{2}}}$$
C.
$${\left[ {\frac{{3T}}{\rho }\left( {\frac{1}{r} - \frac{1}{R}} \right)} \right]^{\frac{1}{2}}}$$
D.
$${\left[ {\frac{{2T}}{\rho }\left( {\frac{1}{r} - \frac{1}{R}} \right)} \right]^{\frac{1}{2}}}$$
When small droplets coelesce to form a bigger drop, energy released in this process is given by,
$$4\pi {R^3}T\left[ {\frac{1}{r} - \frac{1}{R}} \right]$$
According to question
$$\eqalign{
& \frac{1}{2}m{v^2} = 4\pi {R^3}T\left[ {\frac{1}{r} - \frac{1}{R}} \right] \cr
& \Rightarrow \frac{1}{2}\left[ {\frac{4}{3}\pi {R^3}\rho } \right]{v^2} = 4\pi {R^3}T\left[ {\frac{1}{r} - \frac{1}{R}} \right] \cr
& \Rightarrow {V^2} = \frac{{6T}}{\rho }\left[ {\frac{1}{r} - \frac{1}{R}} \right] \Rightarrow V = {\left[ {\frac{{6T}}{\rho }\left( {\frac{1}{r} - \frac{1}{R}} \right)} \right]^{\frac{1}{2}}} \cr} $$
9.
Water is filled in a cylindrical container to a height of $$3m.$$ The ratio of the cross-sectional area of the orifice and the beaker is $$0.1.$$ The square of the speed of the liquid coming out from the orifice is
$$\left( {g = 10\,m/{s^2}} \right)$$
If $${v_2}$$ is the required speed then
$$\eqalign{
& {A_1}{v_1} = {A_2}{v_2} \cr
& {\text{or}}\,\,{v_1} = \frac{{{A_2}}}{{{A_1}}}{v_2} = 0 .1{v_2}\,......\left( {\text{i}} \right) \cr} $$
From Bernoulli’s equation, we have
$${P_a} + \frac{1}{2}\rho v_1^2 + \rho g\left( {3 - 0.525} \right) = {P_a} + \frac{1}{2}\rho v_2^2 + 0\,......\left( {{\text{ii}}} \right)$$
After solving above equations, we get
$$v_2^2 = 50\,{m^2}/{s^2}$$
10.
To break a wire, a force of $${10^6}\,N/{m^2}$$ is required. If the density of the material is $$3 \times {10^3}\,kg/{m^3},$$ then the length of the wire which will break by its own weight will be