CHHANDAKSIR PHYSICES CLASSES
Solution of CSPCJEE/1
1. → = 2𝑖̂ − 3𝑗̂ − 𝑘̂ and → = −6𝑖̂ + 9𝑗̂ + 3𝑘̂ . So → = −3(2𝑖̂ − 3𝑗̂ − 𝑘̂ )
𝐴
𝐵
𝐵
As → = −3 →
𝐵
𝐴
So → || → and → .→ ≠ 0
𝐵
𝐴
𝐴 𝐵
So they are not perpendicular.
2. If the Y axis is divided by mass ,
𝐹(𝑁)
2𝑘𝑔
= a m/s2
The area under acceleration-time graph is the velocity at t when acceleration changed up to t.
𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝐹−𝑡 𝑔𝑟𝑎𝑝ℎ
=velocity
𝑚𝑎𝑠𝑠
So, velocity =
100𝑁𝑠
2𝑘𝑔
= 50𝑚/𝑠
3. Tension is force, and surface tension is force/length. So they do not have the same dimension.
4.Given mass(m)=2kg
Initial radius of the path(r1)=0.8cm
Initial angular velocity(𝜔1)=44rad/sec
Final radius of the path(r2)=1m
Initial moment of inertia,
I1=mr12=1.28kgm2
Final moment of inertia , I2=mr22=2kgm2
From the law of conservation of angular momentum, we get,
I1𝜔1=I2𝜔2
𝜔2=28.16rad/sec.
5.Let the mass per unit length of the rope = 𝜇.
Let a be the acceleration of the rope
So, a=F/m=F/ 𝜇𝐿……..(i)
Now, consider the motion of part Ab which has mass 𝜇𝑥. Therefore the equation of motion will be
F-T=( 𝜇𝑥)a
𝐹
Or, T=F- 𝜇𝑥𝑎 = F= 𝜇𝑥(𝜇𝐿)……..(using (i))
𝑥
𝐿−𝑥
)
𝐿
Or, T=F(1- 𝐿 ) = F(
6.𝜔 =
2𝜋 𝑟𝑎𝑑
60𝑠𝑒𝑐
=
𝜋 𝑟𝑎𝑑
.
30 𝑠𝑒𝑐
𝑅
7.g’ = g( )2
𝑅+ℎ
Given, g’=g/2
𝑅
Or, g/2=g(𝑅+ℎ)2
Or,
1
√2
𝑅
= 𝑅+ℎ
Or, R√2 = (𝑅 + ℎ)
So, h=R(√2 − 1)
2𝜋
8. Maximum velocity vm= a𝜔 = 𝑎( 𝑇 )
So, T=
2𝜋𝑎
𝑣𝑚
= 2𝑥
22 7𝑥10−3
𝑥 44
7
= 10−2 sec = 0.01sec
𝑙
9. In a stationary lift, time period of simple pendulum is given by, T= 2𝜋√𝑔
Where l is the length of the simple pendulum. When a lift accelerates upwards with an acceleration g/4,it
effective acceleration on bob of pendulum, g’ = g+g/4 =5g/4
𝑙
𝑔′
So, T’=2𝜋√
So, T’ =
= 2𝜋√
(
2𝑇
𝑙
5𝑔
)
4
4 𝑙
5 𝑔
= 2𝜋√ ( )
√5
10. In an open organ pipe, both the ends are free ends, hence both ends are displacements antinodes and
hence, pressure nodes.
11. Sound wave is an example of mechanical wave which requires a medium for its propagation. In outer
space, we have a vacuum and hence no sound wave propagation is possible there.
12.Internal energy and entropy are state variables.
13. For 16 g of He, n1=16/4 = 4
For 16 g of O2 , n2 =16/32 = ½
For the mixture of the gas, Cv =
𝑛1 𝐶𝑉1 +𝑛2 𝐶𝑣2
𝑛1 +𝑛2
Where Cv=f/2R,f is the degree of freedom.
Cp =
𝑛1 𝐶𝑝2 +𝑛2 𝐶𝑝2
𝑛1 +𝑛2
where CP =(f/2+1)R
For He, f=3,n1=4 and for O2 , f=5 and n2=1/2
So,
𝐶𝑃
𝐶𝑉
=
5
1 7
2
2 2
3
1 5
(4𝑥 𝑅)+( 𝑥 𝑅)
2
2 2
(4𝑥 𝑅)+( 𝑥 𝑅)
=
47
29
= 1.62
14. Molar heat capacity =molar mass x specific heat capacity
So the molar heat capacities at constant pressure and constant volume will be,28Cp and 28CV respectively.
So, 28CP-28CV=R
CP – CV = R/28
15. According to Stefan’s law, rate of energy radiated by a black body at a absolute temperature T is
given by, H= A 𝜎𝑇 4 , where A is the area of the emitting surface,
T’=T+ 5/100T=21/20T
𝐻′
21
So, = ( )4 = (1.05)4
𝐻
20
% increase in rate of heat energy =
𝐻 ′ −𝐻
𝑥100
𝐻
16. According to Newton’s law of cooling,
60−50
60+50
= 22%
𝜃1 −𝜃2
𝑡
So,
= 𝐾[
− 25]
10
2
1=Kx30
After another 10min,let the temperature be 𝜃,
50−𝜃
50+𝜃
So, 10 = 𝐾 [ 2 − 25]
Solving the equations we get, 𝜃=42.850C
𝜃1 +𝜃2
2
= 𝐾[
− 𝜃0 ]
17. Action of the book on the table is vertically upwards. Therefore angle between the two = 1800
18. According to law of conservation of linear momentum, we get
m1u1+m2x0 = m1v1+m2(-v2) = (m1-m2)v1……………………(i)
𝑢
𝑚 −𝑚
so, 𝑣1 = 1𝑚 2 …………………………………….(ii)
1
1
According to law of conservation of kinetic energy, we get,
1
1
𝑚1 𝑢12 = (𝑚1 + 𝑚2 )𝑣12 ……………………………….(iii)
2
2
𝑢
𝑚1 +𝑚2
…………………….(iv)
𝑚1 −𝑚2
1
𝑚1 −𝑚2 𝑚1 +𝑚2
= 𝑚 −𝑚
𝑚1
1
2
1
3
Divided (iii) by (ii) we get, 𝑣1 =
From (ii) and (iv) we get,
𝑚
On solving ,we get, 𝑚1 =
2
19. Moment of inertia of rod AB and CD about an axis passing through their point of intersection O and
𝑚𝑙 2
perpendicular to the plane of the rods is,I0 = 12 +
By symmetry IXY+IXY’ = Io
𝑚𝑙 2
12
=
𝑚𝑙 2
6
𝑚𝑙 2
6
𝑚𝑙 2
I= 12
Or, 2I =
Or,
20.Her, in this case lens used by person should form image of distance object at a distance of 40cm in
front of it.
u= −∞ , 𝑣 = −40𝑐𝑚.
1
1
1
And 𝑓 = 𝑣 − 𝑢
1
𝑓
1
𝑓
Or, =
Or, =
1
1
−
−40
−∞
1
−40
Of, f= -40cm
100
Power = 𝑓 =
100
−40
= −2.5𝐷
Negative sign shows that lens used in concave lens.
21. When the lens is in air,
1
1
1
= (𝜇𝑔 − 1)(𝑅 − 𝑅 )
𝑓
𝑎
1
20
1
1
1
2
= (1.5 − 1)(𝑅 − 𝑅 )
1
2
When the lens is in water,
𝜇𝑔
1
1
1
= ( − 1) ( − )
𝑓𝑤
𝜇𝑤
𝑅1
𝑅2
Or, fw=78.2cm [putting 𝜇𝑔 = 1.5, 𝜇𝑤 = 1.33]
The change in focal length =78.2-20=58.2cm
22. Given E=4T-T2/200
𝑑𝐸
2𝑇
𝑇
=4−
=4−
𝑑𝑇
200
100
𝑑𝐸
𝑇
𝑛
At neutral temperature, T=Tn, 𝑑𝑇 = 0 = 4 − 100
Or, Tn =4000C,
Also, Tn-Tc=Ti-Tn
Or, Ti=2Tn-Tc
Or, Ti=2 x 400-0=8000C
2𝑥4
23.Total resistance of the circuit =
+ 0.8 =
2+4
Total current in the circuit = 84/12.8A
Reading of ammeter = 8.4/12.8 x 2/6=2.18A
12.8
6
Ω
24.According to Einstein’s photoelectric equation
1
2
hc/𝜆=𝜙 + 2 𝑚𝑣𝑚𝑎𝑥
ℎ𝑐
2( −𝜙)
so, vmax=√ 𝜆
= 8.2𝑥105 𝑚/𝑠
𝑚
25.According to Einstein’s photoelectric equation
h𝛾 = ℎ𝛾0 + 𝐾𝐸𝑚𝑎𝑥
For first experiment , hx4x1015= h𝛾0 + 𝑥
For the second experiment , hx6x1015= h𝛾0 + 3𝑥
From eq.(i0 and (ii) we get, 𝛾0 = 3𝑥1015Hz
26.de Broglie wavelength 𝜆 = 𝑚
ℎ
𝑝 𝑣𝑝
=
ℎ
𝑚𝑒 𝑣𝑒
So, mpvp=meve
𝑣
𝑚
Or, 𝑣𝑝 = 𝑚 𝑒
𝑒
𝐸𝑝
𝐸𝑒
=
𝑝
1
𝑚 𝑣2
2 𝑝 𝑝
1
𝑚 𝑣2
2 𝑒 𝑒
=
𝑚𝑒
𝑚𝑝
<1
So, Ep<Ee
1
27. R= R0( )𝑛 , where n is the number of half lives.
2
Given, R=R0/16
1
Or, R0/16 = R0( )𝑛
2
Or, n=4
Four half lives are equivalent to 8s. Hence ,2s is equal to one half-life. So, in one half-life activity will fall
half of 1600Bq,i.e 800Bq
28. According to Einstein’s photoelectric equation
h𝛾 = ℎ𝛾0 + 𝐾𝐸𝑚𝑎𝑥
so, Kmax= h𝛾 − ℎ𝛾0
K.E
(0,0)
𝜃
𝛾
−ℎ𝛾0
Graph is plotted between K.E vs 𝛾. Compare it with the equation of the straight line , slop of the line =h,
So, slope of the line =Plank’s constant.
29. The energy of an electron in nth orbit of the hydrogen atom is given by,
En=-13.6/n2eV
For thye first excited state, n=2.
So, E1 = -13.6/22 eV
E2 = -13.6/32 [for the second excited state,n=3 ]
So the ratio of E1 and E2 will be= 9/4
30.The magnetism of the magnet is due to the spin motion of electrons. The spinning electrons possess
magnetic dipole moment. This is much greater than that due to orbital motion of electrons around the
nucleus.
3𝛼 (𝐴−12) 5𝛽 (𝐴−12)
𝐴
𝑍𝑋 →
(𝑍−6)𝑌 →
(𝑍−1)𝑌
𝑁𝑜.𝑜𝑓 𝑛𝑒𝑢𝑡𝑟𝑜𝑛
𝐴−12−(𝑍−1)
𝐴−𝑍−11
=
= 𝑍−1
𝑁𝑜.𝑜𝑓 𝑝𝑟𝑜𝑡𝑜𝑛
𝑍−1
31.Let,
So,
𝜀
32.The capacitance of the air filled parallel plate capacitor is given by, 𝐶 = 𝑑0𝐴
When a slab of dielectric constant K and thickness t is introduced in between the plates of the
capacitor,its new capacitance is given by,
𝜀0𝐴
C’=
𝑑−𝑡(1−1/𝐾)
Since a metal sheet of thickness d/2 is introduced hence here, ,t=d/2, K=∞ (for metal)
Or, 1/K=0
𝜀
2𝜀
So, C’= 0𝐴 = 0𝐴
𝑑−𝑑/2
𝑑
Hence, from the above two eq. we get,
2
C’/C = = 2: 1
1
33.All eight charges are symmetrically situated at the right vertices of the cube,each at a distance r equal
to half the longest diagonal from the centre.Length of the longest diagonal of the cube,
d= √𝑏 2 + 𝑏 2 + 𝑏 2 = √3b
√3
hence, r=d/2= b
2
potential at the centre of the cube due to all the ight chatges is,
1
𝑞
4𝑞
V= 8x4𝜋𝜀 . 𝑟 = 3𝜋𝜀 𝑏
√
0
0
34.Serise resistance Rs=R1+R2
𝑅 𝑅
Parallel resistance RP =𝑅 1+𝑅2
𝑅
So, 𝑅𝑆 =
𝑃
Or, √
𝑅1
𝑅2
(𝑅1 +𝑅2 )2
𝑅1 𝑅2
𝑅2
𝑅1
+√
1
=
2
𝑛
1
= √𝑛
35. According toi the Faraday’s laws of electrolysis m=Zit
𝑚
𝐼 𝑡
4𝑥2𝑥60
So, 𝑚1 = 𝐼1 𝑡1 = 6𝑥40
2
2 2
So, m2 =(m1/2)=m/2
[m1=m2]
36. to compute the emf induced consider an element of length dx of rod at a distance x from wire. Emf
𝜇0 𝐼
induced across this element is, de=2𝜋𝑥
𝑋𝑣𝑑𝑥
The emf induced across the ends of the rod due to its motion in the field of the wire, e=∫ 𝑑𝑒 =
𝑎+𝑏 𝜇0 𝐼 𝑑𝑥
𝜇 𝐼
𝑎+𝑏
= 0 𝑙𝑛
∫𝑎
2𝜋
𝑥
2𝜋
𝑎
37.Force experienced by AB and CD are equal and opposite, so the contribution in the net force
experienced by loop is only due to AD and BC. Magnetic field at the location of AD is,
𝜇 𝐼1
B1= 2𝜋(50 𝑐𝑚)
= 4𝑥10-5T⨂
Magnetic force experienced by AD is F1 =IB1 x (15 cm)=3x10-5 N towards wire.
𝜇0 𝐼1
Magnetic field at the location of BC, B2 = 2𝜋(20
= 1x10-5 T⨂
𝑐𝑚)
Magnetic force experienced by BC is, F2 =I2B2 x(15cm)= 0.75x 10-5N away from wire.
Net force experienced by the loop is F=F1-F2 towards wire =2.25x10-5N towards the wire.
38. When the deflection is reduced by 1/n,it means range of the ammeter becomes n times the full scale
deflection current of galvanometer..
So, 5Igx5=IgxG
Or, G=25Ω
If further 1Ω resistance is connected then the situation would be, xX1=25Ig
Or, x=25Ig
So, I=Ig[1+25+5]=31Ig
So, deflection gets reduced to 1/31 of its initial value.
𝜃
𝐵𝐴𝑁
= 160𝑑𝑖𝑣/𝑚𝐴
𝑘
𝐵𝐴𝑁
= 8𝑑𝑖𝑣/𝑚𝑉
𝑘𝑅
39. current sensitivity, Is= 𝐼 =
𝜃
Voltage sensitivity Vs=𝑉 =
For V=0.01mV
𝜃 = 0.08𝑑𝑖𝑣
40.The two bodies will collide at the highest point if both cover the same vertical height in the same time,
So,
𝑣12 𝑠𝑖𝑛2 𝜃
2𝑔
=
𝑣22
𝑔
=
1
2