phy 10th numericals solution

8/11/2019 Phy 10th Numericals Solution

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PHYSICS X

Numerical

Chapter 10

Time period T = 2sec

Length on earth Le = ?

Length on moon Lm = ?

Gravitational force g = 10 m/s2

Solution:

Formula:

By taking square on both side.

Length on moon


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2.

Length l = 1m

Time period T = 4.9sec

G = ?

Solution:

Formula:


3.

Length = 1m

Time period on earth Te =?

Time period on moon Tm = ?

ge = 10 m/s2

gm = 10/6m/s2

Solution:


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Formula:

T(e) = 2sec

On moon

T(e) = 4.9sec

4.

T = 2sec

gm = 10 m/sec2

l = ?

Solution:

Formula:



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5.

Number of waves. N = 100

t = 20sec

Frequency f = ?

t = ?

Length l = 6cm or 0.06m

v = ?

Solution:

Formula:

Time period;

Vilocty

Answers


5/39

F = 5 Hz

T= 0.2 sec

V = .3m/sec

6.

f = 12 Hz

= 3cm or 0.03 mV = ?

Solution:

Formula:

7.

f = 190Hz

l = 90m

t = 0.5 sec

Time period;

Speed;


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8.

F = 4.8Hz

V = ?

t = ?

Solution:

Formula:

Time period;


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9.

Width of ripple tank = d = 80cm or 0.8m

f = 5 Hz

Or T = ?

Solution:

Time period;

10.

f = 90 MHz = 90 106Hz

wavelength = ?

v = 3 108m/s

Solution:

Formula:


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Ch 11

Intensity of normal sound I = 3.0 10-6

Wm-2

Level of sound = 100dB

Intensity level on dB = ?

Sound level = ?

Solution:

Sound level = Sound level =

Sound level =

Sound level = 10 log (310-6 1012)

Sound level = 10 log(3 1012-6)

Sound level = 10 log(3 106)

Sound level = 10(6.4)

Sound level = 64.8 dB

Intensity:

Sound level =


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2.

Sound level = 80dB

Sound intensity = ?

Io = 10-12Wm-2

Solution:

Formula:

Sound level =

3.


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V = 330 m/sec

or 5 10-2mF = ?

Solution:

4.

T = 1min or 60 sec

Heart beats = 72

In 60 second = 72

In 1 second = F = 1.2 Hz

Solution:

Time period;


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5.

V = 1500m/sec

T = 1.5sec or = .75sec (one side time of waves)Distance = ?

Formula:

S = VT

S = 1500 0.75

S = 1125m

6.

T = 5sec or= 2.5 sec (one side time of waves)

S = ?

V = 346 m/sec

Formula

S = VT

S = 346 2.5

S = 865m

7.

T = 3.42sec or = 1.71 sec (one side time of waves)S = ?

V = 531 m/sec

Formula


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S = VT

S = 531 1.71

S = 2618m

8.

F(high) = 20000Hz

F(low) = 20Hz

V = 343 m/sec

?For high level frequency

For low level frequency

9.


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F = 2kHz or 2 103Hz

35 cm or 0.35mS = 1.5km or 1.5 103m

T = ?

V = ?

Solution:

Formula:

Ch 12

1

P = 10CM

Q = -5a

F = ?

Formula:


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2

Ho = 30cm

P = 10.5cm

Q = ?

H1 = ?

F = 16cm

Formula:


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3

P = 20CM

Q = 20cm

F = ?

Formula:

4

P = 34.4CM

Q = -5.66cm

F = ?


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Formula:

5

P = ?

Q = -11.5

F = -13.5

Formula:


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7

R = 38cm F = 19CM

P = 50cm

Q = ?

Formula:

8

Ho = 4cmP = 12cm

Q = ?

H1 = ?

F = 8cm


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Formula:

9

Ho = 10cm

P = 20cm

Q = ?

H1 = ?

F = -15cm

Formula:


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Ch 13

q = 100c = 10010-6

Cnegative charge = 1.6 10-19C

number of points. = ?

Solution:

number of points =

number of points = 6.25 1013

2.

Q1 = 10 c or 1010-6C


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Q2 = 5 c or 510-6C

R = 150cm 1.5m

F = ?

Solution:Formula:

3.

F = 0.8 N

r = 0.1m

q = ?

Solution:

Formula:

As


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By taking square root on both sides.

4.

S =5cmor 5 10-2m

F = 0.1N

F = ?

S = 2cm or 210-2m

Solution:

Formula:

As

To find out the value of force when r = 2cm


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5.

V = 104 V

Q = 100 c or 10010-6 C

W = ?

Solution:

Formula:

W= VQ

W = 104 10010-6

W = 1 j

6.

Q = +2C

VA = 100V

VB = 50V

W = ?

Solution:

Formula:

V = (VA-VB)V = (100- 50) = 50V

W = VQ

W = 50 2

W = 100 j


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7.

V = 9V

Q = 0.06C

C = ?

Solution:

Formula:

Q = CV

8.

V = 6v

Q = 0.03C

C = ?

V = ?

Solution:

Q = CV


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Q = CV

9.

C1 = 6 F or 610-6F

C2 = 12 F or 1210-6F

V = 12VEqualing capacitance. Ce = ?

Q = ?

V1 = ?

V2 = ?

Solution:

Charge


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VOLTAGE

10

C1 = 12 F

C2 = 6 F

Ce = ?

Q1 = ?

Q2 = ?

Solution:

FORMULA

Ce = C1+ C2

= (6+12)

Ce = 18 F

Q1 = C1V

= 12 10-612

= 144 F


26/39

Q2 = C2V

= 6 10-612

= 72 F

Ch 14

Time = 1 min = 60sec

Current = 3mA = 310-3

Formula :

Q = I t

Q = 310-3 60

= 180 10-3C

2

R = 100000

V = 12 v

I = ?

(II)

R =

V = 12

I = ?

Formula :

V= IR


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(II)

FORMULA

V= IR

3

R = M 6V = 100V

I = ?

Formula :

V= IR


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4

V = 10V

I = 1.5A

T = 2min = 120sec

W = ?

Formula

W = VIt

W = 101.5120

W = 1800j

5

R1 =

k

R2 = kRe = ?

I = ?

V1 = ?

V2 = ?

Formula:

Re = R1+R2

= 2+8

= k


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V1 = IR1

= 110-3 2103

= 2V

V2 = IR2

= 110-3 8103

= 8V

6

R1 = kR2 = kV = 6V

Re = ?

I = ?

V1 = ?

V2 = ?

Formula:


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7

P = 100W

V = 220V

R = ?

FORMULA:

P = IV

100 = I220

I = 100/220

I = 0.45A

V = IR

R = V/I

R = 220/0.45

R = `

ENERGY IN KW HOUR= ENERGY IN KW HOUR=

ENERGY IN KW HOUR=


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8

P = 150w

R = Formula:

P = IV or P = I2R

I2 = P/R

I2 = 150/95

I2 = 1.58

I = 1.25A

V = IR

V = 1.25 95

V = 119 VOLT

9

BULB = 100W

HEATER = 4KW

V = 250V

I = ?

R = ?

FORMULA;

BULB:

P = IV

I = P/V


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I = 100/250

I = 0.4A

P = IV or P = I2R

R = P/I2

R = 100/(0.4)2

R = FOR HEATER;

P = IV

I = P/V

I = 4000/250

I = 16A

P = IV or P = I2R

R = P/I2

R = 4000/(16)2

R =

11

R = V = 3v

I = 0.5A

P = ?

P = I2R

P = 0.50.55.6


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P = 1.4w

TOTAL POWER OF BETTRY

P =IV

P = 0.5 3

P = 1.5W

CH 15

VP = 240V

VS = 12V

NP = 2000

NS = ?

FORMULA:

2

VP = 20V


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NP = 100 : 1 =

VS = ?

FORMULA:

3

VP = 170V

N = 100 : 1 =

IP = 1.0Ma

IS = ?

VS = ?

Solution:

FORMULA:


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4

VP = 240V

VS = 12V

NP = 4000

NS = ?

EFFICIENCY = 100%

IP = ?

IS = 0.4A

Solution:

FORMULA:


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5

P = 500 MW = 500106W

V = 250Kw = 250103w

I = ?

Solution:

P = IV

Ch # 18

Half life of N

16

7 = 7.3 sec

Time = 29.9

Remaining quantity of N = ?

Number of half life =

After first half life =


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After 2ndhalf life =

After 3rd

half life =

After 4th

half life =

2

Half life of Co-60 = 5.25 years

Time = 26 years

Remaining quantity of Co = ?

Number of half life =


After 2nd

half life =

After 3rd

half life =

After 4th

half life = After 5thhalf life =

3

Half life of C-14 = 5730 years

Initial quantity = 1

Remaining quantity of C =



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After 3rd

half life =

Total half life for

of C= 3

Total years = 3 5730

= 17190

= 1.72104 years

4

Half life of technetium = 6 hours

Total quantity of technetium = 200 mg

Quantity after 6 half life = ?

Quantity of 6th

half life is =

=

= 3.125 mg

5

Half life of redio active = 10 min

Initial rate = 368 / min

Count rate for 23 = ?

After 10 minuts =


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After 10 minuts =

After 10 minuts =

After 10 minuts =

Total half life = 4

Time = 10 4 = 40 min

9

Half life of C-14 = 5730 years

Initial quantity = 1

Remaining quantity of C =



After 3rd

half life = Total half life for

of C= 3Total years = 3 5730

= 17190

phy 10th numericals solution

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