Senior Secondary Physics (312) Assignment Solution - NIOS

This document presents a detailed solution to a senior secondary physics assignment, likely from the National Institute of Open Schooling (NIOS) curriculum. The solution covers a range of physics concepts, including Newton's second law of motion, Kepler's laws of planetary motion, rotational motion, moment of inertia, angular acceleration, specific heat, and the principles of telescopes. The assignment also addresses concepts related to spring constants and gravitational forces. Each question is solved step-by-step, providing explanations and calculations to arrive at the final answers. The assignment covers a wide array of physics topics, making it a valuable resource for students studying the senior secondary physics course.

Ques. 1
(a)⃗
F=3 i+ 4 j
m=2 kg⃗
a=?⃗
F=m⃗ a
Where F is force on a body
m is mass of body
And a is acceleration
So,
3 i+ 4 j=2⃗ a
1
2 (3 i+4 j)=⃗ a
1.5 i+ 2 j=⃗ a
Magnitude is
a= √ ( 1.5 ) 2+ ( 2 ) 2
a=2.5 m/s2
According to newton’s 2nd law of motion.
The direction is same as direction of force.
So,
tan θ= y
x
tanθ= 2
1.5
tanθ=1.33
θ=tan−1 1.33
θ=53.130 ¿ x−axis
Ques.2
(b)
The planet revolve around the sun which have relation between time period (T) and distance
between sun and planet radius (r) (for circular orbits) is

T 2=r3
Let us consider mass of orbit is m.
And mass of sun is M.
We know that M >>>> m
Radial force is equal to gravitational force
GMm
r2 =m ω2 r
ω2=GM
r3 ………………(1)
T = 2 π
ω
ω= 2 π
T ……………..(2)
From equation (1) and (2)
(2 π )2
T2 = GM
r3
T 2= 4 π2 r3
GM
So we have that
T 2=r3
Distance of Uranus from earth is 20AU
Distance of Pluto from earth is 40AU
So,
TUranus
2
T Pluto
2 = 203
403
TUranus
2
T Pluto
2 = 8
64 = 1
8
TUranus
2
T Pluto
2 = 1
2 √ 2
Ques. 3
(a)
Mass = 5kg

Diameter = 0.4m
r = d/2
r = 0.2m
n1=2rev /sec
The angular speed increases 8 revolutions per second in 2sec.
n2 =8 rev /sec
ω=2 πn
ω1=2 π ×2
ω1=4 π
ω2=2 π × 8
ω2=16 π
Angular acceleration (∝)
∝= ω2 −ω1
t
∝=(16 π−4 π )
2
∝= 2 π (8−2)
2
∝=6 π rad /s2
Moment of inertia of wheel is
I =M r2
I =5 ×0.22
I =0.2 N / m2
Torque ( τ )=I ×∝
τ =0.2 ×6 π
τ =3.77 Nm
Ques. 4
(a)

The specific heat at constant volume (Cv ¿ The amount of heat required to increase in
temperature of 1kg mass by 1K and given volume of mass is constant.
Cv= ( ∆ Q
∆ T )v
The specific heat at constant pressure (C p) it is defined as the amount of heat required to
increase in the temperature of 1kg by 1K and pressure is constant.
C p=( ∆Q
∆ T )p
Cv is given by
Cv= ( ∆ Q
∆ T )v
= f
2 R
Cv= f
2 R
AndC p is given by
C p=Cv+ R
C p= f
2 R+ R
C p=R ( f
2 + 1 )
Cv
Cp
=
f
2 R
R ( f
2 +1 )
cv
c p
=
f
2
f
2 +1
cv
c p
=1+ 2
f
For diatomic gas f =5
γ= Cv
C p
=1+ 2
5

γ= Cv
C p
=5+2
5
γ= Cv
C p
=7
5
γ=1.4
Or
Cv= 1
2 fR
Cv= 5
2 R
C p=Cv+ R
C p= 1
2 fR + R
C p= 5
2 R+ R
C p= 7
2 R
γ= Cv
cp
=7
5
γ=1.4
Ques. 5
(b)
The larger diameter of telescope lens collect more light and therefore more effective for
seeing unclear objects. This is very important because all of the astronomical objects are
generally unclear and we can’t be seen by smaller objective lens. And the larger object lens
effective for resolving deep detailing of objects.
The rays emerging from distance celestial bodies travel by space and atmosphere of earth.
The density of atmosphere is variable. Space is a rarer medium and earth’s atmosphere is a
denser medium. So the ray is enters in dense medium it get deflect. And all rays are subjected
by same phenomenon. What we receive is a bundle parallel rays. Each ray enters the earth’s
atmosphere at different locations that are closed by. So each ray get s deflected along a
different path. Scattering effect takes place. When we observe with our eyes small number of
rays enters our eyes. The image generated is of low pixel value. The reflector is in a positive

to gather more rays and present it to us. As we narrow the area of observations the greater
increases. So we get to see a better image.
Ques.6
(b)
Let us assume the scale can read, M = 50kg
Maximum displacement of the spring = length of scale (l)= 20cm = 0.2m
Time period, T = 0.6s
Force exerted on the spring, F = Mg
Where g is gravitational acceleration
g = 9.81 m/s2
F = 50 × 9.81=490.5 N
Spring constant, k = F
l
k = 490.5
0.2
k= 2452.5
We know that
T =2 π √ m
k
0.6=2 π √ m
2452.5
m= ( 0.6
2 π )
2
×2452.5
m=22.36 kg
Weight of body ¿ mg=22.36 × 9.81
mg=219.35 N