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)
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
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
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
$H=\frac{\rho_1h}{\rho_2-\rho_1}$
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) $\frac{\rho gl h^3}{6}$
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
$\tau=\int_{0}^{h}{\rho gly(h-y)dy}=\frac{\rho gl h^3}{6}$
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 =7.7X103 kg/m3
Density of mercury = 13.6X103 kg/m3
Which one of the following is true
(a) x=32mm and y=28mm
(b) The pressure difference (p2-p1) is 4632 N/m2
(c) The weight of the mercury cube is 16.632 g
(d) None of these
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
$p_2-p_1=\rho_wg(.06-x)+\rho_mgx$
$=1 \times 10^3 \times 10(.06-.032)+13.6 \times 10^3 \times .032 \times 10$
$=280+4352=4632 \ N/m^2$
$W=\rho_iVg = =7.7 \times 10^3 \times (.06)^3 \times 10 =16.632 \ g$
Also Read
