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Multiple Choice Questions
Question 1
A bowl of soup rests on a table in the dining car of a Rajdhani train. The acceleration of the train is g/4 in the forward direction.
What angle does the surface of the soap make with the horizontal?
(a) ${tan}^{-1} \frac {1}{4}$
(b) ${tan}^{-1}{4}$
(c) ${cos}^{-1}{4}$
(d) ${sin}^{-1}{4}$
Solution
In the non-inertial frame of reference of the train (and the soup bowl) , any mass experience a fictitious force of magnitude ma=mg/4 in the direction opposite to the horizontal forward acceleration of the train. The effective gravitational force is the resultant vector
$m\mathbf{g_{\mathrm{efec}}}=m\mathbf{g}-m\mathbf{a}$
The soup is in equilibrium; its surface must be perpendicular to gefec. This implies that the surface makes an angle
$\theta={tan}^{-1}{\frac{1}{4}}$
With the horizontal, with the high side of the surface toward the rear of the train (opposite to the acceleration)
Match the column
Question 2
A block of ice is floating in a liquid of relative density k contained in a beaker
When the ice melts completely, match the column
Column I (k values)
(A) k > 1
(B) k < 1
(C) k=1
Column II
(P) Liquid level will increase
(Q) Liquid level will decrease
(R) Liquid level will remain unchanged
(S) No appropriate match given
Solution
A liquid of density $\rho _L$ contained in beaker filled up to level A
When a block of ice of mass m floats in the liquid, let the level of liquid rise to B
AB is the increase in level of the liquid
If V is the volume of the liquid displaced by the block of ice
Then applying Archimedes principle
Weight of Ice block = Weight of liquid displaced
i.e
$mg=V\rho_Lg$
$V=\frac{m}{\rho_L}$ ---(1)
When the ice melts completely, let the level of liquid + water formed be at C. The volume contained in the height AC is due to the ice converter into the water. Let this volume be V'
Then
$V'=\frac{m}{\rho_w}$ --(2)
From equation (1) and (2), we have
$\frac{V'}{V}=\frac{\rho_L}{\rho_w}=k$
Now for k>1, we have V'> V, so liquid in the beaker will rise
For k< 1, we have V' < V, so liquid in the beaker will fall
For k=1, we have V' = V, so liquid in the beaker will remain unchanged
Linked Type comprehensions
(A)A body of density ρ
_{1} is dropped from rest at a height h into a lake of density ρ
_{2} (ρ
_{2} >ρ
_{1})
Neglect all the dissipative effects
Question 3
Calculate the acceleration of this body while it is in the lake
(a) $g(\frac{\rho_2}{\rho_1}-1)$ Upward direction
(b) $g(\frac{\rho_2}{\rho_1}-1)$ Downward direction
(c) $g(1-\frac{\rho_1}{\rho_2})$ Upward direction
(d) $g(1-\frac{\rho_1}{\rho_2})$ Downward direction
Solution
Let us assume the volume of the body is V
Since $\rho _2 > \rho _1$ Buoyant force of the lake is greater than weight of the body
The body will enter with some speed in the lake and it will be resisted by the buoyant force of the lake.
So Apply Newton’s law (choosing upward direction as positive)
$B-W=ma$ ---(1)
Now
$B= \rho _2 Vg$
$W= \rho _1 Vg$
$m=\rho _1 V$
Substituting these values in equation (1)
$a=g(\frac{\rho_2}{\rho_1}-1)$ Upward direction
Question 4
The maximum depth to which the body sinks before returning to float on the surface is proportional to the height h
And it is given by
H=kh.
What is the value of k
(a)$\frac{\rho_1}{\rho_2-\rho_1}$
(b)$\frac{\rho_2}{\rho_2-\rho_1}$
(c)$\frac{\rho_1}{\rho_2+\rho_1}$
(d) None of these
Solution
Velocity of the body entering the lake can be found using kinematic equation
$v^2=u^2+2as$
Here u=0 s=-h a=-g (taking upward direction as positive)
$v^2=2gh$
or
$v=\sqrt{2gh}$
Now similarly apply this in lake
$u=-\sqrt{2gh}$
$a=g(\frac{\rho_2}{\rho_1}-1)$
v=0
s=-H
$v^2=u^2+2as$
$2gh=2Hg(\frac{\rho_2}{\rho_1}-1)$
Or
$H=\frac{\rho_1h}{\rho_2-\rho_1}$
(B)Water stands at a depth h behind the vertical face of a dam. It exerts a resultant horizontal force on the dam, tending to slide it along its foundation and a torque tending to overturn the dam about the point X. The width of the dam is L.
The density of the water is ρ
Question 5
The resulting horizontal force acting on the dam due to water column
(a) Proportional to h
^{2}
(b) Proportional to density ρ
(c) Proportional to L
(d) None of these
Solution
The pressure a depth y is given by
$P=\rho gy$
We can neglect atmosphere pressure, since it acts on the other side of the dam also
Let us assume a small rectangular strip at a depth y of width dy
The Horizontal force on the strip due to water column
$ dF=\rho gyldy$
So total force on the dam
$F=\int_{0}^{h}dF=\frac{\rho gl h^2}{2}$
So all the three are correct
Question 6 The torque of the forces about point X is given by
(a) $\frac{\rho gl h^3}{3}$
(b) $\frac{\rho g l^2h^3}{6}$
(c) $\frac{\rho gl h^3}{2}$
(d) $\frac{\rho gl h^3}{6}$
Solution
Torque of the force DF about an axis through X is in magnitude
$d\tau=(h-y)dF=\rho gly(h-y)dy$
Total Torque about point X
$\tau=\int_{0}^{h}{\rho gly(h-y)dy}=\frac{\rho gl h^3}{6}$
Multiple Choice Questions
Question 7
A piston of Weight W has the form of circular disk with radius R
_{1} . The disk has a hole into which a thin walled pipe with a radius R
_{2} is inserted. The piston can enter a cylinder tightly and without friction and it is initially at the bottom of the cylinder. M kg of water is poured into the pipe. Find the height of the piston above the bottom
(a) $H=\frac{1}{\pi R_1^2\rho}\left[M+\frac{W}{g}\frac{R_2^2}{R_1^2-R_2^2}\right]$
(b) $H=\frac{1}{\pi R_1^2\rho}\left[W-M\frac{R_2^2}{R_1^2-R_2^2}\right]$
(c) $H=\frac{1}{\pi R_1^2\rho}\left[M-\frac{W}{g}\frac{R_1^2}{R_1^2-R_2^2}\right]$
(d) $H=\frac{1}{\pi R_1^2\rho}\left[M-\frac{W}{g}\frac{R_2^2}{R_1^2-R_2^2}\right]$
Solution
The pressure at the bottom of the cylinder will be given by
$p=\rho g(H+h)$ -- (1)
Pressure at the bottom can be calculated other ways also
$p=\frac{W+mg}{\pi R_1^2} $ --(2)
Equating (1) and (2)
$H+h=\frac{W+Mg}{\rho\pi g R_1^2} $ ---(3)
Now for calculating h, we can see the forces on the piston,
$W=\rho gh\pi(R_1^2-R_2^2)$
Or
$h=\frac{W}{\rho\pi g(R_1^2-R_2^2)}$
Substituting this value in equation (3)
$H=\frac{W+mg}{\rho\pi g R_1^2}-\frac{W}{\rho\pi g(R_1^2-R_2^2)}$
Or
$H=\frac{1}{\rho\pi g}\left[\frac{W+mg}{R_1^2}-\frac{W}{R_1^2-R_2^2}\right]$
$H=\frac{1}{\rho\pi g}\left[\frac{WR_1^2+MgR_1^2-WR_2^2-MgR_2^2-WR_1^2}{R_1^2(R_1^2-R_2^2)}\right]$
$H=\frac{1}{\pi R_1^2\rho}\left[M-\frac{W}{g}\frac{R_2^2}{R_1^2-R_2^2}\right]$
Question 8
A tank contains water on top of mercury. A cube of iron .06 m along each edge is sitting upright in equilibrium in the liquid.
Density of iron =7.7X10
^{3} kg/m
^{3}
Density of mercury = 13.6X10
^{3} kg/m
^{3}
Which one of the following is true
(a) x=32mm and y=28mm
(b) The pressure difference (p
_{2}-p
_{1}) is 4632 N/m
^{2}
(c) The weight of the mercury cube is 16.632 g
(d) None of these
Solution
Now
x+y=.06
so y=.06-x
(.06-x) equals the distance protruding into the water
x equals the distance protruding in mercury.
The net vertical force due to the liquid is $p_2A-p_1A$ where $p_2$ ,$p_1$ are the pressures at the upper and lower face of the block and A is the face area of the block.
For equilibrium, we have
Weight of block= $p_2A-p_1A$
$W=p_2A-p_1A=(p_2-p_1)A$ --(1)
Now
$W=\rho_iVg$
$p_2-p_1=\rho_w g(.06-x)+\rho_m gx$
Substituting the values in equation (1)
$\rho_iVg=\rho_wgA(.06-x)+\rho_mgxA$
Or
$7.7(.06)=(.06-x)+13.6x$
$x=.032m$
So x=32mm and y=28mm
Now
$p_2-p_1=\rho_wg(.06-x)+\rho_mgx$
$=1 \times 10^3 \times 10(.06-.032)+13.6 \times 10^3 \times .032 \times 10$
$=280+4352=4632 \ N/m^2$
$W=\rho_iVg = =7.7 \times 10^3 \times (.06)^3 \times 10 =16.632 \ g$
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