CSIT Family An online CSIT community

### Numerical Problem

52. A force of 5N acts on a body of weight 10N the acceleration of body is 

50 m/s2

2 m/s2

5 m/s2

0.5 m/s2

Solutions
Force = mass * Acceleration
Weight = Mass * Gravity
10 = M * 9.8
Mass = 1 kg
5 = 1 * a
a = 5 m /s2

56.A rain drop of radius 0.3mm has terminal velocity 1m/s in air.The viscosity of air is 18 x 10-5 poise, then viscous force is 

1.01 x 10-7N

1.01 x 10-8N

1.6 x 10-9N

1.6 x 10 -8

Solutions
Radius of drop r=0.3 mm=0.03 cm
Terminal velocity $v_{c}$ =1m/s = 100 cm/s
Viscosity of air η=18×10−3 poise
Viscous force $F=6πηrv_{c}$
​ ∴ F=6π×18×10−3×0.03×100=1.0178 dyne

58.A glass vessel contains air at 27°C the temperature to which it must be heated so that one-fourth of air is expelled from it at constant pressure is 

47°C

77°C

102°C

127°C

Solutions
Let initial number of moles of air at 27°C(300 K) = n
At temperature T , the no. of moles left = n - 1/4 - 3n/4
At constant pressure and volume, n1T2=n2T2
​ n x 300 = 3n/4 x T
300n = 3n
1200 = 3 x T
400 = 3T
127

60. A cylinder of 0.6 m long and 6mm radius is rated at 1.5 kW.the temperature of filament if Stefan’s constant is 6 x 10-8 Wm -2K -4 

920K

1125K

1025K

1320K

Solutions

61. Light of wavelength 550 nm falls normally on a slit of width 22 x 10-7m, the angular position of second minima from central maxima will be 

14.5°

30°

42°

62°

Solutions
$sinθ = {nλ \over{d}}$
Here, n = 2
$sinθ = {2 × 550 × 10^{-9} \over {22 x 10^{-7}}}$
$sinθ = {1\over2}$
$θ = π \over 6$
$θ = 30°$

62. Two lens of power 5D and -7.5D are placed in contact.The equivalent focal length of combination will be 

-40cm

+40cm

-25cm

+25cm

Solutions
$p = {1 \over{ f }}$
${ 1 \over f } = {1\over{f_1}} + {1\over{f_2}}$
${ 1 \over f } = 5 - 7.5$
${ 1 \over f } = 2.5$
f = { 1 \over 2.5 m}
-25cm

63. The length of the microscope is 14 cm and magnifying power is 25 for relaxed eye the focal length of the eyepiece is 5 cm then a distance of object from objective is 

2.4cm

2.1cm

1.5cm

1.8cm

Solutions
$L∞=v_{o}+f_{e}$
$⇒14=v_{o}+5$
$⇒v_{o}=9cm$
Magnifying power of microscope for relaxed eye
$m = {v_o \over u_o}.{D\over f_{e}}$
$25 = {9 \over u_o}.{25\over 5}$
$u_o = {9 \over 5}$
1.8cm

64. A sound wave has frequency 500Hz and velocity 360m/s. What is the distance between 2 particles having phase difference 60° 

0.7cm

1.2cm

70cm

12cm

Solutions
frequency (ν)=500Hz
velocity (v)=360m/s
∴ wavelength = velocity/frequency
λ = 360m/s / 500 = 0.72m
60° out of phase implies that the wavelength between two points is
${x\over60°} = {0.72m \over 360°}$
x = 0.12m
or 12cm

67.A wire of resistance 12Ω is bent to form a circle.The effective resistance between the two points on any diameter of circle is 

12Ω

9Ω

6Ω

3Ω

Solutions
$R_{eq} = {6 x 6 \over 6 + 6}$ = 3Ω

69. A 10 eV electron is circulating in a plane at right angles to a uniform field of magnetic induction 10-4, the orbital radius of the electron is 

18cm

16cm

11cm

9cm

Solutions
Given :
K=10eV
B=10−4Wb/m2
Using : mv=Bqr where mv=√2Km
$r = \sqrt{2Km\over B^2q^2}$
$r = \sqrt{2 × 9.1 × 10^{-31} × 10e \over {(10^{-8})e^2}}$
​ $r = \sqrt{2 × 9.1 × 10^{-31} × 10e \over {(10^{-8})e^2}}$
$1.06 × 10^{-2}m$
11cm

74. A nucleus reactor has a power of 16 kW if the energy per fission is 200 MeV then number of fissions per second is 

5 x 1017

5 x 1016

5 x 1015

5 x 1014

Solutions
Power of nuclear reactor =16KW
Energy per fission =200MeV
$P = n({E\over{t}})$
$16 × 10^3 ={ n × 200 × 10^6 × 1.6 ×10^{-19} \over t}$
${n\over{t}} = 5 × 10^{14}$
(Given 1eV=1.6×10−19J )

56. If a 200m bridge of steel has length from, temperature varies from 243K us 313K, The change of length of bridge for above seasonal variation in temperature (αsteel = 11 × 10-6) 

15.4 cm

15.9 cm

15.8 cm

10 cm

Solutions
The length of steel bridge is: Li = 200m
The initial temperature is: Tt = 243k
The final temperature is: Tf=313k
The coefficient of linear expansion of steel is: (αsteel = 11 × 10-6)
ΔLo=Liαs(Tf−Ti)
ΔLo=200×11×10−5(313−243)
=1.54m or 15.4cm

61. While standing still, an observer measures frequency of tuning fork to be 524 Hz. What frequency does he measure as he approaches the form at Sm/s ? (speed of sound in air = 340 m/s)

500 Hz

420 Hz

520 Hz

532 Hz

65. If critical angle for a material placed in air is 30° the refractive index will be 

2

0.5

1.5

2.5

Solutions
Equation for the critical angle is
$θ_c=sin^{−1}({{n_2}\over{n_1}})$
${{n_2}\over{n_1}}=sin(θ_c)$
${{n_2}\over{n_1}}= 0.5$
since Second medium is air n2 = 1
n1 = n2/0.5
n1 = 2

66. A body has 1c of negative charge. How many excess electron it has as compared to its neutral state? 

6.25 x 1018

6.25 x 10-19

1.6 x 10-19

1.6 x 1018

Solutions
Q = n x e
1 = n x 1.6 x 10-19
n = 1/1.6 x 10-19
n = 1019/1.6
n = ${100\over16}×10^{18}$
n = 6.25 x 1018

71. The self inductance of coil is MH. If a current of 2a is flown. What is the magnetic flux through coil? 

0.02 Wb

0.01 Wb

0.05 Wb

10 Wb

53. A man walks 8 m forwards East and 6m towards North. The magnitude of displacement is 

14m

2m

8m

(82 + 62)m

Solutions
A man walks 8m towards East and then 6m towards North
Using pythagoras theorem P 2 = 82 + 62

54. If initial velocity of a projectile is doubled, the maximum range will by 

2 times

4 times

8 times

16 times

Solutions
$R ={ u^2sin2θ \over g}$
Since angle and G is constant
$R ∝ u^2$
${R' \over R} = {(u')^2 \over u^2}$
$R' = {(2u)^2 \over u^2} R$
$R = 4R$

57. A weightless rubber balloon has 100g of water in it. Its weight in water will be 

100g

50g

> 100g

zero

Solutions

58. The temperature of a substance increase by 270°C. This increase is equal to 

300k

0k

27k

273k

Solutions

63. A person cannot see objects clearly beyond 50 cm. The power of the lens to correct the vision is 

+5D

-5 D

-0.5 D

-2 D

Solutions
$P = 100 \over f$
$P = 100 \over -50$
$P = -2 D$

65. Three capacitors each of capacitance 3pF are connected in series. The net capacitance is 

1μF

(1/3)μF

3μF

9μF

Solutions
${1\over c' }= {1\over c} + {1 \over c} + {1\over c}$
${1\over c' }= {1\over 3} + {1 \over 3} + {1\over 3}$
$3\over 3 = 1$

72. The equation of sound wave is y = 0.0015 sin(62.4x + 316t) the wave length of sound wave is 

0.2 unit

0.1 unit

3.3 unit

not known from the given equation

Solutions
K=2π/wave length
62.4=2π/wave length
62.4=2×3.14/wave length
Wave length=6.24/62.4
=1/10 =0.1m

73. A wave is reflected from a rigid support. The change of phase on reflection will be 

0

π / 2

π / 4

π

Solutions

51. A force F = $\stackrel{\to }{\mathrm{2i}}+\stackrel{\to }{\mathrm{3j}}-\stackrel{\to }{\mathrm{4k}}$ newton acts on a body. In 5 seconds the force displaces the body through a displacement S = $\stackrel{\to }{\mathrm{3i}}+\stackrel{\to }{j}-\stackrel{\to }{\mathrm{2k}}$ meter. The power (in watt) of the force is 

4.3

3.4

5.4

6.3

52. Two simple harmonic waves are represented respectively by y1 = A1 sin ωt and y2 = A2 cos ωt. The phase difference between the two wave is

3π / 2

π

π/2

π/4

53. A river is flowing west to east at a speed of 5 m/min. A man on the south bank of the river, capable of swimming at 10m/min in still water, wants to swim across the river by the shortest path. He should swim in the direction

due north

30° East of North

60° East of North

30° West of North

Solutions

57. At what temperature the rms velocity of a hydrogen molecule be equal to that of oxygen at 47°C 

-273°C

273K

20K

20°C

Solutions

59. A square plate of side 2 units is heated through 10°C. If the linear exansivity of the material is x, increase in the surface area of the plate is 

480x

80x

240x

240x2

Solutions

60. A black body radiates heat energy at the rate of 2 x 10 5Js''m' at a temperature of 127°C. The temperature of the black body at which the rate of heat radiation is 32 x 105 is 

327°

927°

527°

727°

Solutions

64. Two vehicles are approaching each other with a speed of 10ms-1. If A emits a note of 1000 vibrations, what will be the frequency of the note of 1000 vibration, what will be the frequency of the note as heard by a person sitting in B if the velocity of sound is 330 ms-1? 

950Hz

1000 Hz

1062.5 Hz

1215.5 Hz

Solutions

65. An open pipe and a closed pipe resonate to the same tuning fork. The ratio of their lengths is 

1:1

2:1

1:2

4:1

Solutions

67. A 25W, 120V lamp and 100W, 120V lamp are connected in series across a 120V line. Which lamp will burn more brightly 

25W, 120V

100W, 120V

Both will have same brightness

Alternatively both may be bright, one at a time.

Solutions

69. The ratio of induced emf in a coil having 50 turns and rotating with 50 Hz to that of induced emf in a coil having 100 turns and rotating with 150 Hz in the same magnetic field is 

1:2

2:3

1:6

3:2

Solutions

70. A 0 - 100mA resistance 1ft is to be converted to an ammeter in the range of 0-1A. We have to use shunt 

0.1Ω

0.111Ω

0.055Ω

Solutions

71. The ratio of specific charge of proton to that of an alpha particle is 

1:4

1:2

4:1

2:1

Solutions
Let the specific charge of proton be p.
Therefore specific charge of alpha particle = 2p ( ∵α particle has 2 protons and 2 neutrons,
Therefore, ratio of specific charge of protons to alpha particles = 2p/p = 2/1 ⇒1:2

74. The kinetic energy of a particle is doubled, de Broglie wavelength becomes 

2 times

4 times

2 times

1/√2 times

Solutions
$λ ={ h\over p}$
$λ ={ h \over \sqrt{2mE}}$
If kinetic energy is double then λ = $1\over \sqrt{ 2}$

56. The depth of water at which air bubble of radius 0.4mm may remain in equilibrium (surface tension of water = 72 x 10-3 N/m, g = 9.8 m/s2 is given by 

7.348 cm

0.918 cm

3.674 cm

1.837 cm

Solutions
T = 72×10−3 N/m
R = 0.4×10 −3m
Pwater= 103kg/m3
g = 9.8m/s2
Now,
$hdg = {2T \over R}$
$h = {2T \over{ pgR}} = {2 × 72 ×^{-3} \over {10^3 × 9.8 × 0.4 × 10^{-3} }}$

57. When a cylinder is heated, it's length increases by 2%, the area of its base will increase by: 

0.5%

1%

2%

4%

60. The efficiency of Carnot engine working between steam point and ice point is: 

16.8%

26.8%

36.8%

46.8%

Solutions
Temperature of source, T1= 100˚C = 373 K
Temperature of sink, T2= 0˚C= 273 K
$η = 1 - {T_c \over T_h}$
$1 - { 273 \over 373} = 0.268$ 26.8

69. The radius of Earth 6400 km. The capacitance is 

6400 MF

6400 F

1 F

711 μF

Solutions
Radius = 6400km = 6400 x 103m
ε_o = 8.85 x 10-12
$C = 4πε_oR_1$
C = 4 × 3.14 × 8.85 ×. 10--12 × 6400 × 10-4
C = 7.11 x 10-6F
C = 711 μF

70. Two wire of same material have lengths L and 2L and are cross reaction areas 4A and A respectively. The ratio of their specific resistance would be 

1:1

1:8

8:1

1:2

Solutions
The resistivity of a wire is given by $R = {pl\over A}$
$R_1 \over R_2$ = ${pl_1 \over A_1} \over {pl_2 \over A_2}$
Here p is resistivity of material so it will be same
$R_1 \over R_2$ = $l_1 \over A_1$ x $A_2 \over l_2$ $R_1 \over R_2$ = $l \over 4A$ x $A \over 2l$
1:8

74. The half-life of a substance is 1600 years. The mean life is 

1600 yrs

3200 yrs

800 yrs

2309 yrs

Solutions
Mean life = Half Life / 0.693 = 2309 years

55. Two wires A and B are of the same material. Their lengths are in the ratio of 1:2 and diameters are in the ratio of 2:1. If they are pulled by the same force, their increase in length will be in the ratio. 

2:1

1:4

1:8

8:1

Solutions
We know that Young's modulus Y = $2 \over {πr^2}$ x $L\over {l}$
Since Y, F are same for both the wires, we have,
$1\over {r_1}^2$ $L_1 \over l_1$ = $1 \over {{r_2}^2}$$L_2 \over l_2$
$l_1 \over {l_2}$ = ${r_2}^2 × L_1 \over {r_1}^2 × L_2$
$l_1 \over {l_2}$ = ${D_2 \over 2}^2 × L_1 \over {D_1 \over 2}^2× L_2$
$l_1 \over {l_2}$ = ${D_2}^2 × L_1 \over {D_1 }^2× L_2$
$l_1 \over {l_2}$ = ${D_2}^2 \over {2D_2 }^2$ x $L_2 \over 2L_2$ = $1\over 8$
$l_1 : l_2 = 1:8$

56. Water is flowing through a tube of the non-uniform cross-section. If the radius of the tube at the entrance and the exit is 3:2, then the ratio of the velocity of liquid entering and leaving the tube is:

8:27

4:9

9:4

1:1

Solutions
$A_1 V_1 = A_2 V_2$
$π(3)^2.V_1 = π(2)^2.V_2$
${V_1 \over V_2 }= {4 \over 9}$

58. 50g of ice at -6°C is dropped into the water at 0°C. how many grams of water freeze? Sp. Heat capacity of ice=2000JKg-1C-1.

1.785 g

4.25 g

3.16 g

9.33 g

60. The temperature of source and sink of carnot engine are 400k and 300k respectively. What is it's efficiency?

100%

75%

33.33%

25%

Solutions
It is given that T = 400 K and t = 300 K
These are the temperatures of source and sink respectively
The efficiency of the cycle is,
$η=1−{t\over{T}}$
η=1−0.75=0.25
Thus the efficiency is 25 %

61. The refractive index of air with respect to glass is 2/3. The refractive index of diamond with respect to air is 12/5. Then the refractive index of glass with respect to diamond will be. 

5/8

8/9

5/18

18/5

Solutions
$μ_{air} \over μ_{glass}$ = $2\over3$
$μ_{diamond} \over μ_{air}$ = $12\over5$
on multiply we get
$μ_{diamond} \over μ_{glass}$ = $8\over5$
$μ_{glass} \over μ_{diamond}$ = $5\over8$

63. Light of wave length 7200 Å in air has wave length in glass equal to ......... (given Refractive index of glass with respect to air = 1.5)

7200 Å

4800 Å

1080 Å

10800 Å

Solutions
Wavelength in medium = R.I.ofmedium / wavelengthinair
7200 Å / 1.5
4800 ​

64. Three capacitor of capacitance 3µF, 9µF and 18µF are connected first in series and then I parallel. the ratio of C equivalent capacitance in two cases (CI/CP) will be

1:15

15:1

1:1

1:3

Solutions
$1 \over{C_s}$ = $1 \over 3$ + $1 \over 9$ + $1 \over 18$
$1 \over{C_s}$ = $1 \over 2$
${C_s}$ = 2μF
${C_p}$ = 3 + 9 + 18 = 30 μF
${C_s}\over{C_p}$ = $2\over30$ = $1\over 15$

66. The resistance of two lamp are in ratio of 1:2. Their wattage will be in the ratio of

1:2

2:1

4:1

1:4

67. The energy stored in 50mH inductor carrying a current of 4A is .......

0.1 J

0.4 J

0.3 J

0.01 J

Solutions
L = 0.05
I = 4A
$E = {1\over 2 }LI^2$
E = 0.4 J

69. A particle of mass 10-31 kg is moving with a speed of 105m/s. The De-Broglie wavelength of the particle is ...... 

6.63 x 10-8 m

6.63 Å

66.3 Å

6.63 x 10-7m

Solutions
$λ = {h \over mv}$
$λ = {6.63 × 10^{-34} \over {10^{-31} × 10^5}}$
λ = 6.63 x 10-8 m

73. A common emitter transistor amplifier has a current fain of 50. If the load resistance is 4 kΩ and the input resistance is 500 Ω. The voltage gain on the amplifier is ....... 

400

6.2

500

300

Solutions
β=50
RL =4kΩ
Ri =500Ω
The resistance gain is given by,
$R_L \over R_{i}$ = $4000 \over 500$ = 8
Now, the voltage gain is
$A_v = β$($R_L \over R_{i}$)
$A_v$ = 50 x 8 = 400

74. A tube closed at one end has a resonating length 1 meter. The air column in the pipe can resonate for sound of frequency ..... (V = 320ms-1)

166 Hz

249 Hz

575 Hz

80 Hz

Solutions
$v ={ 320 \over 4 × 1 }Hz$ = 80Hz

75. A displacement wave is represented by y = 0.025 sin(5006 - 0.0025x), where y,t and x are in cm, second and meter respectively. The wave length of wave is .......

20π m

40 π m

60 π m

80 π m I there is