2018-10-01_Physics_For_You

10. (a) : Let x 1 = a sin Zt and x 2 = a sin(Zt + G) be
    two S.H.M.
    a ata at
    33

= sinaωωnd− =+sin( δ)

sinaωωtt=+^1 ndsin( δ)=−

3

1

3

Eliminating t,c^1 os sin

3

1 1

9

1

3

δδ+− =−

9 cos^2 G + 2cos G – 7 = 0

cosG = –1 or

7

8

i.e., G = 180° or cos−^1 ^7 

9

Now v 1 = a Z cos Zt and v 2 = a Z cos^ (Zt + G)

If we put G = 180°

We nd that v 1 and v 2 are of opposite signs.

Hence G = 180° is not applicable.

∴=δ cos−^1 ^7 

9

11. (c) : ∆φ π ππ
    λ

=⇒ 2 +=θπ

2

nd^2 sin 2 n

2 2 1

2

π

λ

dnsinθπ=− 

sinθ λλ

λ

=−  =×

×

2 1 =

22

1

223

1

12

n

d

⇒=y

()100

1

λ^212

144 100

100

12

25

3

yy^22 =≈()λ ; λλ=

12. (a) : r=°cos3 0 =^23
    2
    T 2 cos 30° + T 1 cos30° + mg cos 30° = mv
    r

2

T 2 sin 30° + mg sin 30° = T 1 sin30°

Just slack Ÿ T 0 = 0° Ÿ T 1 = mg

2 mg cos 30° × L cos30° = mv^2

vL^22

4

=^3 × 10

v^2

2

=^3 ×× 10 24. = 36 Ÿ v = 6 m s–1

13. (b) : Acceleration of electron = ae

? a

eE

e me

==

Force on electron

Mass of electron

Similarly, acceleration of proton, a

eE

p mp

=

Su=+tat

1

2

2

? Sa=+=+ (^01) eptS at
2

0 1

(^12)
2
2
and^2

? 1

2

1

(^12)
2
2
at at^2
ep=
or
t
t
a
a
eE
m
m
eE
m
m
e
pe
pp
e
2
1

^2







==×=?

t

t

m

m

p

e

2

1

12

=









/

14. (b) : VAB = I 2 (4 + 6)
    ? VAB = 10 I 2

5 :

4 :

A B

I

I 1

I 2 6 :

Also VAB = I 1 (5)

? 5 I 1 = 10 I 2

or I 1 = 2 I 2 ...(i)

Let HI==R

Heatgenerated

second

2

?

H

H

I

I

4

5

2

2

1

2

4

5

=

×

×

HI

I

42

2

2

10 2

4

25

=

×

()×

or H 4 10 4 1

45

= × 2

×

= cals−.

15. (a) : Since the electron moves undeected between
    the plates of condenser, there should be equal and
    opposite magnetic and electric forces upon it.
    Electric force is to the le, due to positively charged
    plate.
    Magnetic force on electron should be towards right.
    e magnetic eld should therefore be directed
    perpendicular to plane of paper inwards, according
    to Fleming’s le hand rule.
    Magnetic force = Electrical force
    evB = eE

or B

E

v

V

dv

==









1

B=

×







 ×







− 

3

500

310

1

(^3610)
or B = 0.055 T

16. (a) : e magnetic force on the charged particles
    provides necessary centripetal force for circular
    motion of the particles.
    mv
    r

qvB

2

= or r

mv

qB

=

r

p

qB

r

qB

==or

2 mK (∵ p (^2) = 2mK)
rrpd :: rα= ::

1

1

2

1

4

2

or rp : rd : rD = 1 : 2 : 1 or rD = rp < rd

17. (d) : I = I 0 (1 – e–t/W) where I

V

R

L

(^0) R
==andτ

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