2018-10-01_Physics_For_You

SOLUTIONS

1. When the wire is subjected to repeated alternating
   strains, the strength of its material decreases and the
   wire breaks.


2. The boiling point of a liquid increases with
   pressure. For example, if the pressure is more than
   the atmospheric pressure, water boils at a temperature
   higher than 100° C.


3. Graph (a) represents a brittle material as it indicates
   a very small plastic range of extension.


4. Terminal velocity, v v r^2

? v

v

r

r

A

B

A

B

=









=







=

(^22)
1
4

11 6:

5. In a capillary tube, a liquid rises to a height h given

by h

rg

=









2 σθ

ρ

cos

For water, q is positive and hence h is positive. So water

rises in the capillary tube. For mercury q is obtuse,

cos q is negative and hence h is negative. So mercury

gets depressed in the capillary tube.

6. (a) When two wires of same size are suspended in
   parallel, a force F equal to the breaking force will act
   on each wire if a breaking force of 2F is applied on the
   parallel combination.

(b) F

YA l

l

Yrl

l

==

∆∆.π^2

i.e., F v r^2

Thus for a single wire of double the thickness, the

breaking force will be 4F.

7. Let A be the area of cross-section of the string in SI
   units.
   Mass of string, m = Volume × Density
   = (A × 10) × 1.5 × 10^3 = 1.5 × 10^4 A kg
   Stress in the string

==

mg ××× =× −

A

A

A

15. (^1041015) .N 1052 m

Y

L

=×=

Stress ×

Strain

or

.

/

510

15 10

5

8

5

∆

? (^) ∆L= ××
×

=×− =

1510 5

510

15 10 15

5

8

. .m (^3) .mm

8. Using continuity equation, A 1 v 1 = A 2 v 2

v

v

A

A

r

r

r

r

1

2

2

1

2

2

1

2

2

1

2

4

1

===







=

π

π

9. Volume of the ball, V

m

=

ρ

Mass of the liquid displaced, mV′= =ρ m

ρ

00 .ρ

When the body falls with a constant velocity,

Viscous force = Effective weight of the ball

F = Weight of the ball – Upthrust

= mg – mcg

F = mg –

m

gmg

ρ

ρ

ρ

ρ

(^00) ..=− 1






10. Y Fl
    Al

=

∆

∵ Y, l and A constants,

?

F

∆l

= constant or 'l v F

? If tension is T 1 , then

l 1 – l v T 1 ...(i)

If tension is T 2 , then

l 2 – l v T 2 ...(ii)

Dividing eqn. (i) by eqn. (ii), we get

ll

ll

T

T

(^1) lT lT lT lT
2
1
2
12 2211

−

−

=−or =−

or l(T 1 – T 2 ) = l 2 T 1 – l 1 T 2

or l lT lT

TT

lT lT

TT

=

−

−

=

−

−

21 12

12

12 21

21

11. As the temperature increases, the interatomic
    forces of attraction become weaker. For given stress,
    a larger strain or deformation is produced at a higher
    temperature. Hence the modulus of elasticity (stress/
    strain) decreases with the increase of temperature.
    OR
    Radius of each small drop = r

Volume of each small drop =

4

3

pr^3

Volume of bigger drop =

4

3

pR^3

R^3 = nr^3 Ÿ R = nr

1

3

As terminal velocity, v v r^2

v

v

r

′ R

=







2

, where v' is the terminal velocity of bigger

drop.

v

v

n

′

=

− 2

(^3) ⇒vn′= v
2
3
Since v' = 4v,? 4
2
vn=^3 v or n==()48
3
2