Gravitation Numericals with Solutions for Class 9

a. This can be calculated using the formula
$Relative \; density \; of \; solid = \frac {Weight \; of \; solid \; in \; air }{ Loss \; of \; weight \; of \; solid \; in \; water}$v $= \frac {200}{40} = 5$

b. This can be calculated using the formula
$Relative \; density \; of \; liquid = \frac {Loss \; of \; weight \; of \; solid \; in \;liquid}{Loss \; of \; weight \; of \; solid \; in \; water} = \frac {30}{40} = .75$

The force acting on an object perpendicular to the surface is called thrust

Thrust =150 N
Now $Pressure = \frac {thrust}{Surface \; Area} = \frac {150 }{15 \times 10^{-4}} = 10^5 \; N/m^2$

$1000 \; kg/m^3 =1 \; g/cm^3$

$Relative \; density = \frac {density \; of \; the gold}{density \; of \; water}$
or
$density \; of \; gold = (Relative \; density) \times (density \; of \; water) = 19.3 g/cc$

$Density = \frac {mass}{volume}$
$Density = \frac {350}{200 }=1.75 g/cm^3$
Density in SI unit
$=1.75 \times 1000= 1750 kg/m^3$

$Relative \; density = \frac {density \; of \; the silver}{density \; of \; water}$
or
$density \; of \; silver= (Relative \; density) \times (density \; of \; water) = 10800 kg/m^3$

Thrust= Force=250 N
$Pressure = \frac {Thrust}{area} = \frac {250}{15 \times 10^{-4}} = 166666.7 Pa$

Thrust= 400 N
$Pressure = \frac {Thrust}{Surface area} = \frac {400}{ 25 \times 10^{-4}} = 160000 Pa$
Now New pressure
$Pressure = \frac {Thrust}{Surface area} = \frac {600}{ 25 \times 10^{-4}} = 240000 Pa$
Change in Pressure = 240000- 160000=80000 Pa

Relative Density is the ratio between density of a substance and density of water is called relative density.Relative density has no units because it is the ratio between the similar physical quantities
$Relative \; density = \frac {density \; of \; the object}{density \; of \; water}$

$Relative \; density of mercury= \frac {density \; of \; the mercury}{density \; of \; water}$
or
$density \; of \; mercury= (Relative \; density) \times (density \; of \; water) = 13600 kg/m^3$

Density of hollow ball = 15/20 = .75 gm/cc
Density of solid ball = 8 gm/cc
Now Density of water is 1 gm/cc

So Hollow ball will float and Solid ball will sink

Volume of the dining room= $50 \;m \times 15 \; m \times 3.5 \; m= 2625 m^3$
Now Density of air= 1.30kg/m³
Now
$density = \frac {mass}{volume}$
or $mass = density \times volume= 1.30 \times 2625=3412.5 kg$

Thrust =100 N
Pressure = 100/25 =4 Pascal
New Pressure =25/25 =1 Pascal
Change in Pressure = -3

For solid ball as u=0
$h_1 =\frac {1}{2}gt_1^2$
For hollow ball as u=0
$h_2 =\frac {1}{2}gt_2^2$
From above two, we can find
$\frac {t_1}{t_2} = \sqrt {\frac {h_1}{h_2}}$
Ratio will not change in either case because acceleration remains the same.
In case of free-fall acceleration does not depend upon mass and size.

Method -1
a. Let V be the volume of iron and $\rho _i$ be the density of Iron,$\rho _o$ be the density of oil

Upthrust in Water = 44.5 - 39.5 = 5 gf =.049 N
Now
$ V \times 1000 \times 9.8 = .049$
or $V= .000005 m^3 = 5 cm^3$
Now mass of Iron=44.5 gm
Density of Iron = 44.5/5 = 8.9 g/cc
Now relative density of the iron = Density of iron /Density of watter = 8.9/1 = 8.9

b. Now Upthrust in oil = 44.5 - 40.3= 3.2 gf= .03136 N
Now
$V \times \rho _o \times 9.8 = .03136$
or $ \rho _o= 640 kg/m^3$

Method-2
a. This can be calculated using the formula
$Relative \; density \; of \; solid = \frac {Weight \; of \; solid \; in \; air }{ Loss \; of \; weight \; of \; solid \; in \; water}$
$= \frac {44.5}{5} = 8.9$

b. This can be calculated using the formula
$Relative \; density \; of \; oil = \frac {Loss \; of \; weight \; of \; solid \; in \;oil}{Loss \; of \; weight \; of \; solid \; in \; water} = \frac {3.2}{5} = .64$
Density of Oil =.64 X 1000 = 640 kg/cubic mv

a. Upthrust = Weight in air - Weight in water = 50 -44 = 6 gf = .0588 N
b. Let V be the volume
$Upthrust= V \rho_w g = V \times 1000 \times 9.8 = 9800V$
Now
$9800V=.0588$
or $V= .000006 m^3 = 6 cm^3$
c. Mass of Solid = 50g
$Volume= 6 cm^3$
$Density =\frac {mass}{volume} = 8.33 gm/cm^3$
Now $\text {relative density of the solid} = \frac {\text{Density of Solid}}{\text{Density of water}} = \frac {8.33}{1} = 8.33$

i.Here v=0,u=?, s=10 m ,a= -g= -9.8 m/s²
$v^2= u^2 + 2a s$
$0 = u^2 + 2 \times (-9.8 ) \times 10$v u = 14 m/s
(ii) $v = u + a t$
$0 = 14 - 9.8 \times t$
t = 1.43 s.

Mass of the wooden block = m= 10 kg
Dimensions = $40 \; cm \times 25 \; cm \times 10 \; cm$
Thrust on the table top is due to the weight of the wooden block.Now,
Weight of the box = $F = mg= 10 kg \times 9.8 m/s^2= 98 N$
a. Area of a side = $length \times breadth = 10 \times 25 = 250 cm^2 = 0.025 m^2$
$ Pressure = \frac {weight}{ area}= \frac {98}{0.025}= 3920 N/m^2$
b. Area= $length \times breadth= 40 \times 20= 800= 0.08 m^2$
$ Pressure = \frac {weight}{ area}= \frac {98 }{0.08}= 1225 N/m^2$

$P=\frac {F}{A}$
Now
$P_1= \frac {F}{A_1}$
and
$P_2= \frac {F}{A_2}$
Now
$P_1 : P_2 = A_2 : A_1 = 150 :200= 3:4$

Let two forces $F_1=5x,F_2=7x$
Let two areas $A_1=4y, A_2=5y$
$P=\frac {F}{A}$
Now
$P_1= \frac {F_1}{A_1}$
and
$P_2= \frac {F_2}{A_2}$
Now
$P_1 : P_2 = F_1 A_2 : F_2 A_1 = 25xy : 28 xy=25 :28$

The mass of the earth, M = $6 \times 10^{24} kg$
The mass of the moon,m = $7.4 \times 10^{22} kg$
The distance between the earth and the moon,d = $3.84 \times 10^{5} km$
$= 3.84 \times 10^5 \times 1000 m$
$= 3.84 \times 10^8 m$
$G = 6.7 \times 10^{-11} N m^2 kg^{-2}$
Now
$F= G \frac {M \times m}{d^2}$v Substituting the values
$F=2.01 \times 10^{20} N$

$g= \frac {GM}{r^2}$
Now
$g_s=\frac {GM}{R^2}$
$g_h=\frac {GM}{(R+H)^2}$
Now
$g_h= \frac {1}{2}g_s$
$\frac {GM}{(R+H)^2} = \frac {1}{2} \times \frac {GM}{R^2}$
$(R^2 + H^2 + 2RH)=2R^2$
$H^2 +2RH -R^2=0$
or
$H= \frac {-2R \pm \sqrt {4R^2 +4R^2}}{2}$
or
$H=R(\sqrt {2} -1)$ or $-R( \sqrt {2} +1)$
Ignorning negative.
$H=R(\sqrt {2} -1)$

We know that
$F=\frac {Gm_1 m_2}{r^2}$
(i) 2F
(ii) F/4
(iii) 4F
(iv) 4F
(v) F/2
(vi) 16F

Radius of the orbit of the satellite, r = 44400 km = 444 * 10⁵ Mass of the earth, Me =$6 \times\ {10}^{24}$ kg Universal gravitational constant, G = $6 \times\ {10}^{24}$ kg Now, the acceleration of the satellite, $a = \frac {G \times M_e }{ r^2}$ Substituting the above values $a= 0.203 m/s^2$ .

Given,
time taken=0.5s
g=10m/s²
u=0 as free fall
(i)
u=0 a=g From first law of motion
v=u+at
v=0+10*0.5
v=5m/s
(ii) Average speed = $\frac {u+v}{2}=\frac {0+5}{2}=2.5$ m/s
(iii) From newton's equation of motion, As u=0,
$h=ut+ \frac {1}{2}gt^2$
$h= \frac {1}{2}gt^2= .5 \times 10 \times (0.5)^2 =1.25$ m