hints & solutions mathematics€¦ · ^ bansal classes physics [12] 20 20 (b) wor done by 1k0 v...

H I N T S & S O L U T I O N S MATHEMATICS

1. L = Lim x->0 1

/n(l + x) / n ( x + A / [ + x )

L = Lim ^ ( x W l + x 2 ) - / n ( x t l ) _ = ^ In x + Vl + x 2

1+^ x-,0 x . M ^ > . / n ( x + V T ^ ) x-Zn^x + V T ^ 2 - l ) + l)(x + Vl + x 2 -1)

X ( x + v r + x 2 - i )

note that Lim x->0

/nf(x + Vl + x 2 -1) + 1 X + Vl + x 2 - 1

->1

/Y /n

hence L = Lim x->0

vv x + Vl + x 2

1 + X \ \

- 1 +1 X + Vl + X 2 - l

x + Vl + x 2

1 + x - 1 (x + Vl + x 2 - 1 )

/n (V / — r

x + Vl + x

Note that Lim x->0

vv 1 + x - 1 \ A

+ 1 / y

x + Vl + x" ~ T + x ~

> =i - l

= Lim x + Vl + x 2 - 1 - x = Lim VT + x z - 1

L - Lim

x(l + x)(x + V1 + x 2 -1) x - > 0 x(V 1 + x 2 + x -1)

( V ^ + l - l X V x ^ l + l ) x->0 ( V x 2 + l + l ) - x ( V x ^ + l + x - l )

1 [ ( x 2 + 1 ) - 1 ] [ V X 2 + T - ( X - 1 ) ] = — Lim 2 x - ( V x 2 + l + x - l H V ^ T - - ( x - l ) ]

= Lim x-2 Lim x 1 x-*o 2 ( x 2 + l ) - ( x - l ) 2 2x 2

(as Lim(l + x) = 1) x-M)

••• L = 2 L + 153 (1/2)+ 153 hence — ; = — 7 7 ^ — = 1 + 2 • 153 = 1 + 306 = 307 Ans. ] L (1/4 J

<§Bansal Classes PHYSICS IW

2. Let C: problem is solved correctly P(C) =

W: problem is solved wrongly P(W)

9-10 + 10-16 _ 5 ~ 6 300 ....(1)

6 Cj: She thinks that the problem is correct Wj: She thinks that the problems is wrong

1 A : Problems from the university A; P(A) = —

B : Problems from the university B; P(B) 3 1 P(W/Wj) - P(Wj/A) 1 4 ' 5 '

we have to find P(W/C) = ? P( Wj n C) P(Wj)-P(C/WJ)

now P(W,/C) = " ^ c P = P ( C T now P(Wj) = P(A n Wj) + P(B n Wj)

= P(A) • P(Wj/A) + P(B) • P(W,/B) I _L _ A

~ 3 ' 5 + 3 ' 10 ~ 15 Hence from (1) and (2)

(2/15)(1/4)

3

P ( C / W j ) = 7 ;

....(2)

P(W,/B) 10

A V w , 9 P(W,/C)

ljiF 2Q , 10 V n d l - A W v — |H-q +

_ _ J_ 6 _ _1_ (5/6) ~ 30 5 ~ 25

PHYSICS

p + q = l + 2 5 = 26 Ans. ]

l.(a) 2 ]uF ^ L H f i - H ^ where Q = 40 pC, the charge on 2pF just before closing the switch 2. (Q-q)

By kirchhoffs law (Q-q) I 2 J 1 0 - ^ =21 = 2 , 2 dt

20+ 1 0 - 2 I - q = 0

dq 3q - 20 = -1 dt '

dq = 1 3 q - 2 0 4

qr__dt _ I f 1 integrating both sides J 3 q _ 2 o ~ 4 J = T 3 q - 2 0 • 2 0

20 on solving q = — ( l - e ~ 3 t / 4 } i C dq I = -7- = 5e- 3 t / 4 amp dt

^ Bansal Classes PHYSICS [12]

20 20 (b) Work done by 10 V battery = — x 10~6 x io = — x 10"5 J

2(a) y =

Heat lost by 2 0 resistor = j l 2 ( 2 ) d t = 50 j e~ 3 / 2 d t= — xlO" 5 J o o

percentage of work done by 10V battery lost in the form of heat :

mAD nXD

rio xlO" \

-5

u ^20 xlO"

\ .5 xlO" /

x l0%= 50%]

(b)

d d 8100 m = 4500 n _n _ 9 m 5

9 x 4 5 0 0 x l 0 ~ 1 0 x l

for coincidence of maxima

2x10 - 3 = 2.025 mm

f 0 f 0 f 0 x8100 = f 0 V 2) l 2)

x 4500: ' 0 n + — v 2 j ' r m + —

9 5 => n = 4; m = 2

y = 4 + — V 2y

4500 xlO" 1 0

2x10 13— = 1.0125 mm Ans. ]

mg A The whole process is isobaric with pressure = P 0 +

AQ = nC pAT 7 25 1000 = 1 x - x — x AT 2 3

6000 240

„ ^ 2 5 240 2000 (a) AW = nRAT= 1 x — x — J =285.7 J

10 3 +- 2x10 10x10"' N/m 2 = 1,2x 105 N/m 2

PAY = AW P x A x Ah = AW 50 Ah = —r m 21

(b) Final temperature = (300 + (240/7)) K 4.

2340 K] If x is the position of centre of mass below the centre of sphere

R m xm 2 7 8m ~T

0 X — 14

Total energy = mg

^Bansal Classes

Q \ R - — cose 14 + - I P © 2 ....(1)

PHYSICS [13]

5.

6.

From parallel axis theorem

I p * I 0 - m

Also I 0 = —

R 14

r + m y "8m ̂ ^ 7 j

13R 14 (0 is small);

m v ' A z J

differentiating equation (1) R d0 1 0 = - mg — sin 6 ) — + —

d© _ mgR dt ~ 14I n ]

— = t a n 9 D Ax= dcosO (10 4 X-X)= \04X cos 0

1

2© d© lr dt J ± p

d cos 0

cos 0 = 1 10' ,4 >

2 sin2 (0/2) = - T ( cos 0 = 1 - 2 sin2 (0/2) ); sin (0/2) =

• h y 0 100 D '

<r \ r i ^

r = R sin 0 • i •• . i i

4tcR z J V i

D\fl y = - j ^ - m = 2 cm Ans. ]

°yx(27cRsine)Rde = dE

V

P o s i n 0 d 0 ^ he — = (dn)— 2 1 X

dn = c w > f :

P 0 sin 0 d 0 2hc

h

dF cos 0

' P A s i n G d e ^ dF cos 0 = , x 2hc C O S 0

a(2itRt) I t / 2

JdFeff = — { s i n e COS 0 d 0 2c P 0 1 p = -

e f f 2c 2 4c 0

o2 Tt Rt = 4c P 0 = 8 71 a Rt c


fx 200 340 + 5 + 5 340+ 5-To

350x200 335 A - 200 340 + 5 + 15

340 + 5 - 1 0 200x360

335

f2 = 200 x 360 335

3 4 0 - 5 - 5 3 4 0 - 5 - 1 5

360 330 200 x x 335 320

360 330 350 y 2 - / i = 200 x — x — - 2 0 0 x

200 335 320

/ 2 - / , « 12.68 Hz ] m m u cos 0 = — v 2 + mv.

V 2 ~ V i O-ucos0

v 2 - V t = u cos 0

335 335

u sin 8

360x330 320 -350 200

335 360x33

32 -350

u sin 0

Vj = u COS 0

4ucos 0 = 3

u cos 0 V,

§ = tan~ f ucosO V 3usin0 $ = tan~ cot0

n 2

it

cot 0

- tan" cot0 0

cot (p+0)

tan" cotG n - ( P - 0 )

cot 0 cot P cot 0 - 1 3 "" cot p +cot0

cot P cot 0 + cot 2 e - 3 cot P cot e -3 cot p i

2 tan 6 + 3 tan 0

=> cot p is mm. at tan 0 = j j hence p is max, at 0 = 30° ]

9, V. = AXp V, kX, p _ _ — L p

A 2

11Bansal Classes

AX, i. kX,

1 A i mole vacuum

of diatomic

IP

PHYSICS [15]

AW = JkXdx = Y ( X ^ - X F ) x,

AQ - AU + AW = nC v dT + AW= - R ( T 2 - T , ) + AW

AQ = | ( R T 2 - R T , ) + A W = | ( P 2 V 2 - P 1 V i ) + AW = | ( k X 2 - k X f ) + | ( X * - X ? )

AW (k /2 ) (X 2 - X 2 ) (5 / 2)(kX 2 - kXf) + (k / 2)(X 2 - X f )

100 2 7 ^ x 1 0 0 = — % ]

10, t u u u i v 0 c o s 3 7 ° ' t 2 ev 0 cos37° 5 / 8 x v Q x 4 / 5

A + i L - I i T = t i + t 2 = v 0 v 0 v 0

2v Q sin 37° 13 2 v Q x 3 / 5 _ 13 v A ' 10 v '0

i 13 v 0 = 5 J j m/s

o TTTTTTTTT7TTT7

4 7R 3 11.(a) M = p - n —

F = Fj - F 2

M, tota! j 7

(2R y f—T I 2 J

M = removed 4 R 3

P — 71 = 3 8 . M

" 7 2GMm

9 R 2 ( m 2GMm

9 R 2 \ R 7

(b) If V, and V 2 gravitational potentials, v is final speed

where V G M M a l

2R GM removed

3R 2

G M tola! R

G M y v

removed

R / 2

putting the value of V, and V 2 and solving we get, I 6 G M v = n 1 \ 21R J

12.

Lj r, l 2 r .

O -e = 15 sin 200 t

R, = 3fi a-3 Lj = 20 x 10" H R2 = 6Q L 2 = 40 x 10"3 H

Z , = y 'Rf-HtoLj) 2 = V 3 T 7 4 f = 5 fa Bansal Classes PHYSICS

Z , = , / R 2 + ( c d L 2 ) 2 a/6 2 +8' 10

Z = ^ ( R 1 + R 2 ) 2 + c o ( L 1 + L 2 ) 2 = V 9 2 +12 2 = 15

I - ^-sin(200t-<j)) = j|sin(200t-<|>) R , + R 2 9 3 I = sin (200t - <()) where cos<() = = — = -

Now, V, = IqZ j sin(200 t - <j> + ()>,)

= 5 sin (200t) [ v <|> = (f>,] here cos<j)j = V, = 5 sin(200t) V 2 = I 0 Z 2 sin(200t) V 2 = 10sin(200t) ]

13.(a) dN dt : q - AN A = ln2

N JXT 1 dN - f 4&*-N-q 0

In 2

'dt

l n (AN-q) - ln 4qT ' in 2 = t (-A)

In AN-q 3q -At

(b)

AN 1 q — = - + N = f (1- 3 e-^) 3q 3 X qT N = -r~z (1 + 3e" x t) In 2

At t = 2T

N = ^ - ( l + 3 e - 2 S n 2 ) 2T In 2 qT In 2 1 + V2

No. o f a particle decaying = N 0 + q ( 2 T ) - N 2 T '

"3A/2-3 = qT V2in2 + 2 ]

[N 0 is N at t

fe Bansal Classes Simple Harmonic Motion [5]

1 4 . ( a ) A x 1 2 = D

A(|)

y d _ 0 . 7 x l 0 ~ 3 x 0 . 7 x ! 0

271 2 x 0 . 3

1 2 7 x 1 0 ~ 7 7 x 1 0

71 Similarly, A(j)23 = 2tc + —

x A x 271

_ x 7 x l 0 7 X — = — = 0 n + — " 7 0 . 6 3 3

(c) A x = ( 1 . 2 5 - 1 ) x 1 . 4 x 1 0 - 6 = - ^ x l O " 6 = X 2 A(J) = 71

71 = > A(j)]2 = 7 T + —

1 5 . (a)

A(J)2, = 7 1 +

3

71

1 A ; = A 2 + 4 A 2 - 2A 2 x - = 4 A 2 = 4 I 0

4 1 4 I e„ = 7 x 7 x 1 0 " 4 = — x W / m 2 res (y 3

V x = 1 2 + 1 0 = 2 2 m / s v y = 16 m / s

2 2 V +v x y

V = V256 + 484 = V 7 4 0 m/s

v = V 7 4 0 m / s

2 v (b) T i m e o f flight is = — -§

x = 3 . 2 x 2 2 = 7 0 . 4 m

2x16 10

3 . 2 s e c

1 7 0 . 4 = 1 0 x 3 . 2 + — x a x ( 3 . 2 ) 2

7 0 . 4 = 3 2 + 5 . 1 2 a

a = 7.5 m / s 2

& Bansal Classes PHYSICS [18]

5 j e e p = 7 . 5 i , a b a l l = - 1 0 j

^bal l / jeep = a b a l l ~ a j e e p = - 1 0 j - 7 5 i , bal l / jeep 12.5 m/s 2

' ba l l / j eep ~ 1 2 i + 1 6 j ^ b a l l / j e e p = 2 0 m / s

0 = ( 2 0 ) 2 - 2 x 12.5 x s , s ^>s = 16ffi I

400 25

16. (a) (2FX, - F X 2 ) - i k(X, - X 2 ) 2 - i mvf + i 4mv

(b)

F+ma,

F + t - x m 5m 6F 5

6F

• 2F 2 F - F F

c m 5m 5m

F - 4ma„

2 F - 4 m x -

6F 5

> 6F 5

5m

tm 6F 5

v 2

~ X m , + y X m , = ^ ( X m , + X m . ) 2

fiF 1 — (Xm, + Xm 2 ) = - k(Xm. + Xm 2 ) 2 , 6F k y = - ( X m j + X m 2 ) 12 F 12 5mg 12mg Xm, + Xm, = — ~ = — x —-— = ———

1 1 5 k 5 k k

maximum elongation^ 1 2 m g

(c) mXnij = 4mXm 2, 1 2 m g Xm, + Xm 2 = ^

Xm, = 4Xm,

Xm, = 4Xm 2

1 2 m g 4Xm 2 + Xm 2 = K 1 2 m g 5Xm 0 =

2 k 12mg X m 2 = y r

X m ' = 4 X 5k 12mg 48mg

5k

amplitude of mass m 48mg ~~5k~


a = K _ 5k m x 4 m " f e

5m 5k 48mg , „„ of m = coA = J - — x — - — = g — x24

m a x V 4m 5k s V 5k

17.(a)By momentum conservation, 5m x v 2 - mv, = mvQ ...(1) By angular momentum conservation,

mv 0 — z I 2 J = - mvj — z I 2 J 5 m/

+ ~rT x ® - (2 )

By coefficient of restitution equation, (I x

3 4

V, + VT + CD Z 1 2 U ,

v 0 - Q •••(3)

V0-+V, = 5V2

5v. - z 5/2co 12

^ 6V 7-V 0+CD / ^ — z \ 2 7v„ 6x 5OD/; 12 + © - z

/ — z 2

co/ ( B ) VB = v 2 - Y - 0 => co/

V 2 = T /2co

12 co/ T

I I z =

(C) CO = - VN X -4 0 6 3 t j l + t 4 ' 3 + 9

21VQ 38/

co/ 21v, v2 2

1 K y = 2 x m x

76 ' 29v (

"l6

29v f

v i = 5 v 2 - v 0 = - ^ -

+ —x5mx 2 2i^o

76 + —x5mx — x 2 12


1 2 = ~ m v n 2 0

^ 29 2 + 5 x 2 1 2 + 5 x 7 2

76 2

AK 2845 K„ 5776 xlOO =43%

1 2 3291 „ m v x

2 0 5776

J

18. (a) C v = m p,C + p 2 C v

p , + p 2 2 = 3

'V, 5R c _ 3R 2 ' v2 ~ 2

C = 2 2 15R „ — - + 3R 21R

(b)

C p = R + C v - 2 . 1 R + R = 3.1 R

= S i = M I L _ 31_ 31_ y r 2.1R 2 1 ' Y ~ V

given: T, - 300 K P=const

21

> T 2 = 1200 K . v=c«"t > T 3 = 300 K

(c)

Pj = P 0 2 x 105 N/m 2

V, =Vn

i =const T, = T 4 = 300 K P.j = P 4 = P 0 = 2 x 105 N/m 2

P 2 = 2 x 10 s N/'m2 = P 0

V =4V 2

P Po

Zo 4

300K 1200K" r ' K j

Work done in entire cycle W, l->2 3P 0 V 0

W2-»3 = 0

p P , = f jL =

4 v = 4 V 3 H v 0 ,T=300K

2T=1200K

-sT=3G0K.

m V 4Vn

W3_,J - nRT, log c ZL

V V 3 P 0 V 0 l o 8 e

'•"V ' v o v4Vo J

L 4 P 0 V 0

Q,n = Q i|Bansal Classes

W n e t = w l - > 2 + W2-*3 + W3~>l W n e t - 3 P o V 0 - l - 4 P 0 V 0 = 1 . 6 P 0 V 0

W n e i = I - 6 P 0 V 0

^ 2 = AU ! _ > 2 + W ! ^ 2 = pC p (T 2 -T 1 ) PHYSICS

Qin = Ql-»2 = H X 3 - ! R ( T 2 - T l ) = 3- 1 (H R T 2~ l ^ R T l ) = 3 - l ( P 2 V 2 " P l V l ) Qin = Q i - > 2 = 3-1 (4P 0 V 0 -P 0 V 0 ) = 3P 0 V 0 X 3 . 1

A W net Q V

x lOO

1.6P nV n

ri % - ° ° * 1 0 0 = ——— xlOO = —• x 1 OO 1 1 / 0 3P„V„x3.1 3 , i x 3 9.3 •o'o • Tj %

160 9.3 = 17.2% ]

19.

(a) (b)

20.

1 10~3 sec A T 1000

Imax = 4A2 I 0 = 2A2

when A r e s < J 2 A then detector becomes idle

from 1 to 2, it remains idle for phase angle of 90° => t i d l e = — = 0.5 x 10' 3 s ] A T 2

4.8m, 0.06 kg 2.56m, 0.2 kg P 0.06 Q 0 2 T = 8 0 N R

= i o = 0.0125 kg /m p , 0.0781 kg /m 4.8 A- = 3.5 cm = 3.5 x lQ- 2 m

2.56

f 80 80 V2 V 0.0781 « 32 m/s

A r = A i

V 2 ~ V 1 V v i + v 2 ; 3.5 x 10~2

3 2 - 8 0

V l + V 2 , = 3.5 x 10-2

v 112 , 48 3

— x 3 . 5 x 1 0 - 2 = - - x 10~ 2 m I 1 A* Ja

2 x IO" 7 - m

4.8 2.56

21. Given: I, - 0.5 A I = I 1 + I 2

Applying Kirchoffs Law:-15 - 111, - ( 1 1 - r ) x 0.5 = 0 and IjT = I 2 x 6 = 3 Putting I, - 3/r in eq. (1)

9 . 5 - 1 1 x (3/r) + 0.5r = 0 r 2 + 19r - 66 = 0 r = 3

So, length AD = 3 cm] ^Bansal Classes

•(1) .(2)

PHYSICS

15 V in 1 •A/WV*] I, r n (10—r) -WWV S w v W W B

L . I 2 —'VWv en

[22]

22. 4 3 v = — Ttr dv . dr — = 4 TO-2 — dt dt ttR 4 4T x — 8 T)/ r

| r 3 dr TR 8r)/

4 [l - [dt 7 j

TR 8rj/ 1

t , = 2r)/r t, = T- Ans. i TR

23. 3L/5

\ L

VQCOSS

•V0sin6

o

< ,v0cos8 »v0sin8

Lab frame Fig. (A)

In CM frame velocity of B

vncos8

CM frame Fig. (B)

Tangential velocity remains unchanged whereas velocity change along string for B is vsin9

Impulse of tension : mv sin 9

2

24.(a) Let wooden log moves distance x Displacement of centre of mass along horizontal = 0

2m V

L x 2 , m L x 2 - mx = 0

3L x — — 20.25 cm 8

(b) By conservation of mechanical energy

(m + m + 2m)v 2 = (2m - m)g(Lsin60°);

43.Bansal Classes PHYSICS [23]

25.

point where a bat is held \

From linear momentum linear impulse equation, we get FAt = M V c m ...(1)

From angular impulse angular momentum equation, we get F(y - a)At = (MKL)co ...(2)

( y - a ) (Mk2co) "MVT

For point to be stationary Vcm = ( a - X ) ®

V k 2 k 2 cm = (a - x) = ~~—~ or co ( y - a ) ( y - a ) mg+ 1.36 = p,vg mg + 0.82 = p 2 vg mg + 1 = p vg

P l v l + P 2 V 2 V I + V 2

V1 solving we get 7 Ans. ] 2 Z

27. If system is balance then d i d 2 — = T 0 x — 2 2 2 V t = T 2 * V > T i d i = T 2 d 2 (i)

Mg Mg 2Tj = Mg sin 30° = , T, = ....(ii) T 2 = 10 x 10= 100 N , T 2 = 100 N (iii) from (i) d 2 = 3d, , T, x d, = T 2 x 3d , , T, = 300 N

M x g from (ii) 300 = — , M = 120 kg M = 120 kg ]

28. Ray 1 carries (1 - k) 2 of the beam's energy; Ray 2 carries k 2 ( l - k) 2 ofthe beam's energy; Ray 3 carries k 4 ( l - k) 2 of the beam's energy; etc. The total fraction of transmitted energy is

(1 - k) 2 + k 2 ( l - k ) 2 + k 4(I - k ) 2 + ... = (1 - k) 2(l + k 2 + k 4 + ...) = (1 - k) 2/(l - k 2 ) = (1 - k)/(l + k) = 7/13 = 53.8% ]

^Bansal Classes PHYSICS

29. RB = —— sin 9 RA = RB cos 2 0

h = —-r (cos 26) sin 0 path difference Ap = RB - RA

h Ap = ——— (1 - cos 20) = 2h sin 0 = X sin 0 X 7t 2A, 2 sin 0 = 7- => 3 x — = h 180 2(4m)

71 X = 4x—-m =0.42 ml 30 J

30. Vj - v 2 (35-20) (in meter) 2g 100 Also, v, (1) = v 2 (2) (by equation of continunity)

.2 3v 15 2g 100

.'. v 2 = 1 m/s mass flow rate = a x v 2 x ^ = 1000 x l x 2 x i o - 4

= 0.2 kg /sec

Now, 2 2

V 2 ~ V 3

2g 37.5-35

100 1 - V 3 2.5 100 x20 V 3 = ]

Unl v 31. ^-r suppose v is the constant velocity then co = — m g

t = IAB e 1 1

R 2 Brv

1 = R

t = IAB =

A 1 n 2 V = — x — Br x — = Brv 2R

2R x %r2B I B V v 2 R

current I: when v = 3 m/sec e 1 Brv 1 = 1 0.5x0.2x3 — x 2 0.15 1A Ans R 2 R

for highest velocity net torque on the ring should be zero

x = mgr = 1 B2to*3V -; v = 2mgR 2x20xl0~ 3 x 10x0.15 B V 0.25x7tx4xl0 , - 2 5x3.14 r =0.2 m/sec Ans.

and t = mgr = 20 x 10~3 x 10 x 0.2 - 4 x 1Q-2 N-m Ans. ]


N

1 dN 3 2 " ( a ) N ^ = C v 2 '

IdN v0

V = ICv2dv C v q ^ 1 = 0. C = - J 1 v 0

(b) v„ JvdN v0

= _o = fCv 3 dv = N n 0.75VA

(c)

33. i

j v 2 d N

N = JCv^dv = J ^ x ^ " = v 0 J =0.775 v 0 J

mu = 4mv 2 cosQ ...(1) mvj = 4mv 2 sin9 ...(2) From eq. (1) & (2) u 2 + v? = 16 v 2 ...(3) Ej = -(13.6 eV)z 2 = -54.4 eV

(13.6z) E, = - - 1 = -6.4 eV i n

o -mu

Before collision

mvj 0

v4invo After

collision

34.

AEj = E 2 - Ej = 40.8 eV AE2 = E 3 - Ej = 48 eV AE3 = E 4 - E , = 51 eV AE4 = E 3 - E 2 = 7.2 eV

65eV= ~mvf + | ( 4 m ) v 2 + AE = ~ m v f + ~(4M) u a + v *

16 + AE ...(4)

1 2 1 2 65eV 65eV = - m v . + - m v , + ——- + AE 2 1 8 1 4 On substituting AE 3 = 51 eV mv, comes out to be negative, which implies that, electron transition upto n = 3 is possible. In subsequent de-excitation the possible energies are AEj - 4 0 . 8 eV [n = 2 to n = 1] AE2 = 48 eV [n = 3 t o n = l ] AE4 = 7.2 eV [n = 3 to n = 2] ]

Net external force = M „ a system cm

2mg sin 6 = 2m ( ma + 0 I 2m ,

=> a = 2g sin 6 ^Bansal Classes PHYSICS [26]

m + m . . a = gsinG m

Also v 2 - — = 2ad; 3v_ 8a solving we get d = 2 m/s Ans. ]

35. Let at any instant t charge is q . dq

- 1 = ~ dF [charge decreasing]

di dt

d^q d t 2

di C L dt C dt 2 '

r q = q 0 sin 1 t + (j)

.VLC at t = 0, q = q 0 => <|) = rc/2

f 1 ^ •'• ^ q o c o s ^ t j Now, when energy becomes half then,

2 c 4 c Let it happens at t = t 0

q0 f i

q0

•c '-nmp-

d 2 q d t 2

q LC

V ^ = q ° c o s I V T c ° J ;

t 0 = ^VLC = 5TI x 10-5 sec]

VLC l0 = 7t/4

36.

Just before collision

m. v

m, 2v

L/6 | C L/6i

Just after collision

"co0

some time after collision

Applying conservation of angular momentum about CM ofthe system which is CM ofthe rod just after collision

fL l f L̂ m2v ~ - m v — - m v UJ 2mL

12 •m

2 FLL

— + M —

U J J con

3v 4L


Applying conservation of angular momentum about CM ofthe system just after collision and a the time when insects are at distance x from centre

r 2mL 7 , + mx +mx v 12

co 2mL2

co = 2co0L j L + 2x"

de= 2 L a ) ° d t

- + 2x 2

6

L 2 c d q l / 2 f d e = ^ r J 9V J dx

L/6 12

\ '

+ x

0 r co 0 LVn

9v tan" 1 V 3 - t a n - 1

V3 co0L7r 71 9V3v ~ 12V3 r a d i a n A n s " ]

37. (a) / M g N

v ^ J (R sin 60°) = x = l a

(R sin 60°) r M ^ .'. a v ^ j

- m R 2 + 'm R 2 2 _

a =

a A

gV3 4R

V P v 4 R y

R = gV3

(b) Nj = cos 60° = ~ ~ 1 2 4

mg . m and sm 60° - N 2 = y a A

N = 2 8

hor

= mg + N, cos 60° + N 2 cos 30° 21

= Nj sin 60° - N 2 sin 30° _ mgV3

16 ]

N a \ a

/ ' 6 0 ( \\a \ A A

mg T"

N, '60° 30°

kN,

"hor

mg

^Bansal Classes PHYSICS [28]

38. Time taken by A to reach to B t, = 1/4 now velocity of B will be 4 m/s

time taken by B to oscillate = n -

t 2 = n V I6n' TC

4tc I 4

time taken by A to go back to left wall is = 1/4 — t̂ 1 total time period of this oscillatory system = t, +1 2 +1, = — +

39. tan0

v m < | + , rolling will not occus

mg(sin0~pcos0) m

Also a pmgR cos 9 pg cos 9 mR 2 R

If t is the time to complete one revolution 1 , — a t 2 = 2p

Distance travelled by centre 1 „ 27ta _ 27tR(sin9-pcos9) _ 2TI(5-V3)

a pcos0 V3 x a t 2 =

mg _ mg 0 \ 40,(a) Initial acceleration = — - - = g/5

(b) At the instant when spring makes an angle 9 with vertical

m/4

x = L tan9 mg xLtan0 - xLtan0 + W = m + — jv 4 s 2 V 4 J

i j

W g = work done by spring force -• - change in spring energy mgLtan© mgLtan© 1, r T T , 2 5 -> — _ - k [ L s e c 0 - L ] = - m v " 2 4 2 8

Bansal Classes PHYSICS

(c) For v = 0 x = V3L => 9 = 60° V 3 m g L _ J _ k L 2

41.

k =

4 2 V3mg

Zm I

J U

h

2L v

| a 0 = 2m/s 2

w.r.t. to lift acceleration of coin, a' = g - a 0 = 8 m/s 2

u r d = 3 m/s u 2 , maximum height, H' = —^ = 0.56 m < 1 m

s 2a' so, it cannot touch the roof of lift. Now, let t 0 is time taken to reach the floor

• - l = 3 t 0 - - ~ ( 8 ) t 2

On solving, t 0 = 1 sec w.r.t. ground

Initial velocity, v 0 = 8 m/s v 2

Maximum height, H = — = 3.2 m Time taken for upward motion, t, = v f f/g = 0.8 s Time taken for downward motion, t 2 = 1 - 0.8 sec = 0.2 Distance travelled during downward motion = (l/2)g * (0.2)2 = 0.2 m Total distance travelled = 3.2 + 0.2 = 3.4 m Displacement till that time = 8(1) - (l/2)g x ( l ) 2 = 3 m ]

1 42. Point source - spherical wavefront => I o c ~ j I, = 41

L = 10 log 1 0~~ = 30 (1)

41 Lj = 10 l og 1 0 ^" = 10[log 1 04] + L = 20 log i 0 2 + 30 « 36 dB sound will be inaudible if I 2 = I 0

from (1) I = 1000IG

I d 2

~ = 1 0 0 0 I 2 ~ 1 W U ~ d ( 2 0 ) 2

d 2 = 2007l0 m ]


43. m = / 40 f + u 40+ (-30)

m = v/u v = -120 cm

v i = m 2 v 0 < ,

64 cm/s

2 cm

2 cm 4 cm/s

Vj = 1 6 x 4

A _ l - ± v u / 1 1 1 1 1 v - 3 0 +40 40 30

120 40 / = 40 cm / = 60 cm

v = -120 = 64 cm/s

Generaly equation for y-component of velocity of image yi / y0 / + x o ' y«

f s

f + x Oj

dy 0 y 0 f dx0 u x m U _ x u dt v x 0 x y of = mv.. _ — - x (1) ( / + x 0 ) 2 dt " " " v o - ( f + x 0 ) 2

V Applying equation (1) for first lens with v Y o = 0 and then applying for second lens with v y o = - 16/5 cm

v = °y _

y of dx o

( / + X 0 ) 2 2dt = - (16 /5 ) cm/s

Now substitute Vq and m = 60

3 _ _ 5 X

y

/ 1 zr\ • 1 6

60+ (-160) 64x2x60

in eq. (1), to get

\ -> j (60-160) 144/125

Therefore, relative velocity = 2 x 144 125

44.(a) To determine the specific heat capacity of unknown solid,

we use s m,s 3 + m 2 s 2

solid m, t i

~0SS J and get s S 0 l ] d = 1/2 cal/g/°C

fds^ = 2A0 V s ) max

1 1 • + -e s s - e 2 o , - o 2 (0.1 °C) 1

- + -

1 ss y 40.0-20.0 80.0-40.0 1%

(b) s. = m 2 s 2 + m 3 s 3

m. v 9 j - e y y substituting value, we get s, = 0.5 cal/g°C for finding error in s.

r \ m 2 s 2 + m 3 s 3

as, = m i y ( e a - e ) ( d e - d e 2 )

( G , - e ) 2 ( 0 - e , ) ( d 0 , - d 0 )

^Bansal Classes PHYSICS [31]

ds^ _ (de - de 2 )(6 1 - 9) - (9 - e 2 ) (dO, - d6) (6! - 9) + (9 - 9 2 ) de - de 2 (0, - 0) - d0j (0 - 6 ̂ ) S , ( 0 1 _ 0 ) ( 0 _ 0 2 )

As, _ ( 0 1 - 0 2 ) A 0 + ( 0 1 - 0 ) A 0 2 + ( 0 - 0 2 ) A 0 1

M ( H ) ;

As, is minimum when (0,-9) (9 - 0 2 ) is maximum.

(0,-0)(0-02)

As, _ A9[2(9 , -9 2 ) ] s, ( 0 - 0 ) ( 0 - 0 )

This happens when 0 0 , + 0 2

20 + 80 => steady state temperature should be — - — = 50°C Ans. ]

45. 1 cm, 1 cm 46. (a) 0.2 cm; (b) -0.1 cm ; (c) 12 cm [Sol.(c) u = 30.2 - 0.2 (excess reading)

= 30.2 cm v = 19.9 - (-0.1) (excess reading) = 20.0 cm. 1 1 1 7 = - + - =>f= 12.0cm ] f v u

47, t = t Let charge on outer shell is Q dQ

charge on innershell = Q, + Q. - Q => Leakage current I at distance r from centre J = ctE

dt

I Am-'

= s rQi + Qi-Q"

4n £n Kr 2

I <?(Ql+Q2-Q)

e0K

dQ = cr{Qx+Q2-Q) dt ' eo K

I dQ „ ^ a + o7 - o

Q=Q2 1 * 2 ~ f=0

t-t - f SL-dt

-in Qx + Q2-Q a

Qi Gn k Qx + Qi-Q

Qi

at e„k

48. The expression for the electric field can be obtained as E = k (1 + cos Qt) cos cot

1 1 - k cos cot + — k cos (co - Q)t + — k cos (co + Q) t,

Q = Q 2 + Q , at e„k


When the product of the two cosines is expanded using a standard trigonometric identity. These three terms corresponds to photons of energies hco, h (co-Q). The latter exceeds the ionization energy by 0.7 eV. That difference equals the ejected electron energy. ]

49. One can use the formula for the capacitance of a parallel-plate capacitor: C = e 0 A/d. According to the definition of capacitance, the charge on one plate is Q = CE, where E is the emf (and the voltage across the capacitor). Also, the current is given by I (t) = dQ / dt I (t) = E (d C / dt) 1 (t) = E (dC/dd) (dd/dt) One can see that (dC/dd) = ( -s 0 A/d2) Since d « d 0, (dC/dd) « ( - s 0 A / d 2 ) Also, (dd/dt) = (aco sin cot) I (t) = (Ee0A/ d^) (aco sin cot) The amplitude of the current is then I = Es 0A aoo/dg, so

A 1 a = EAWE '

50. The fact that the pendula hang vertically at equilibrium implies that the rubber band tension is zero at that moment. If the bobs are each displaced a distance x, then the rubber band is stretched by 2x, resulting in a tension of 2kx. Gravity and the string tension act together to produce an additional restoring force of mgx/L on each bob. The total restoring force on each bob is then -(2k + mg/L)x when the bobs are moved outward from their equilibrium positions and -(mg/'L)x when they are moved inward.

In general, the period of a simple harmonic motion is T = 2TT/CO. Then, for the first (outward) half-cycle this becomes P, = 27t[m/(2k + mg/L)]1 / 2 = 27t(2k/m + g/'Lf 1 / 2 and for the second (inward) half-cycle it is P 2 = 27i[m/(mg/L)]!/2 = 2?c(g/L)~1/2. To find the period ofthe motion we add the two half-periods: T = (P, + P2)/2 = 7i[(2k/m + g/L)~m + (g/L)""2]

51. In the center of mass (c.m.) reference frame, all the kinetic energy is conserved to electrostatic potential energy at the final instant. The center of mass velocity is 3v/5 (assuming the alpha particle is travelling in the positive direction), so the incoming alpha velocity is 2v/5 and the incoming proton velocity is -8v/5. Therefore, the total kinetic energy in the c.m. reference frame is 8mv2/5. At the final instant, the electrostatic potential energy ofthe system is e2/27ts0r. Solving for r gives r = 5e2/167rs0mv2. ]

fe Bansal Classes Simple Harmonic Motion [5]

52.

53.

54.

Since the bullet's loss of speed (Av = v - u) is small compared to u, we can assume that its deceleration is fairly uniform while passing through the block, so the upward force exerted by the bullet on the block is also fairly steady. The time that it spends passing through the block can be found:

At = d/v a v = 2d/(v + u) The upward impulse exerted on the block is equal to the bullet's loss in momentum:

FAt = m(v - u) so, F = m(v - u)/At. Since the block does not quite lose contact with its support, we set that force equal to the weight of the block : Mg = mAv/At. Solving for M and replacing the Av and At with the expressions above, we find that the minimum block mass is M = m(v 2 - u2)/(2gd) = 6.0 kg ] Minimum 0 implies minimum V and maximum V & .

In order to have the aforementioned situation, the rock has to be launched horizontally. ( g > 0 , V =0) Then: V f y = V 2 + 2gh

V f i ! = v tan 0 = V f /V., so fy fx ' V. = VtanO

fy

V 2 = v 2 tan 2 0 Also: V 2 = 2gh and h - vtair0/(2g,) ] The rate at which heat is transferred by a single rod is given by the expression:

Q , . A - T C P = Y = k A = C ( t h - t c ) in which A is the area, L is the length and k is the thermal conductivity, the only factor different in the two experiments. Since A ^ - t,)/L is a constant, let C denotes it. When you put two rods in series you would have:

p _ A ( t h - t c )

k, k„ For our case, L, = L 2 L. So we have:

A ( t h ~ t c ) L l / k , + l / k 2 y

k,k"

k, + k 7 I V I 2 / _1 P

1 1 — + 1 2

We have T ] = Q/P,, T, = Q/P2 and T = Q/P Q Q Q T = ^ = —+ — = T + T, P P. P„ 1 - 80 min.]

55. First, one can connect the voltage source, the unknown resistor r and the ammeter in series. The voltmeter should be connected in parallel with the ammeter. The ratio of the readings, (V/I) equals the resistance of the ammeter. Then the voltmeter should be reconnected in parallel with both the ammeter and the unknown resistor. The ratio of the new readings (V7T) now equals the total resistance of the ammeter and the unknown resistor. The unknown resistance r is, therefore, given by r = (V'/I') - (V/I).]


56. Since the normal force on the top mass is zero and the horizontal acceleration of the top mass is zero at the instant it loses contact with the wall, the tension in the rod at that moment must be zero. Thus, free-body diagrams for the two masses are as sketched below.

X TRMG Therefore, the top mass has downward velocity v = -dy/dt and acceleration g = -d 2y/dt 2, while the bottom mass has rightward velocity u = dx/dt and zero acceleration. But y = V r 2 - x 2 since the length of the rod is fixd, and thus

v = - dy/dt = i r 2 _ x 2 dt dx xu

y . 2 2

and g - - d 2y/dt 2 u dx xu dy _ u xuv _ y 2 u x~u 2 _ U i y dt y 2 dt

Finally, from conservation of mechanical energy, 1 mg(r - y) = - m(u2 + v 2)

- + + -y

- , . 2 ( ,.2 A

2g(r - y) = u 1 + x 2 2 u r gy

so that 2

y = T r u : |8gr 27

57. The heating element will consist of several segments of wire connected in parallel. For maximum heating power, each segment must carry the greatest possible current, which is 2 A. Therefore, the resistance of each segment must be 110 V/2 A = 55 Q. Since, 536/55 = 9.74, we can only use nine segments. Each segment will have length = (55/536) L, where L = the original length of the wire. (The tenth piece will be too short and must be discarded) The heating power will then be 9 x 110 V x 2.0 A = 1980 W]

58. Let the mass have descended a distance y and be at speed v. The changing magnetic flux through the circuit loop leads to a Faraday emf BLv. This emf is related to the charge on the capacitor as

q = CBLv Differentiating with respect to time gives the current in the loop

dv I = CBL dt The magnetic force on the current carrying bar (upward ifthe bar is descending) is given by

dv F m a o = BIL = C B 2 L 2 — mag d t

The net downward force on the bar and Newton's second law of motion gives

mg - ky - CB 2 L 2 dv dt m-d 2 y

d t 2


mg Transforming to the new variable u = y - — - allows the equation of motion to be written in K.

the form d2u + = 0 d t z m + CB L

This is the familiar simple harmonic motion equation with angular frequency co = and so the period of oscillation is V m + CB 2 L 2

T = 2iZi m + CB 2 L 2 ]

59. That charge is finite because the second inductor shorts out the battery so that the final voltage between the top and bottom of the diagram will be zero. It will take some time to reach that condition because the current in L 2 will approach its asymptotic value gradually. Let "I" be the current through L, and let "i" be the current through L 2 . (Both currents are functions of time.) The voltages across the three vertical parts of the network must be equal:

V RI + LjdI/dt E - r(I + i)

V = L 2 di/dt

...(1)

...(3) Combining eqs. (1) and (3),

RI + L, di/dt L 2 di/dt

...(2)

...(4) It seems likely that V will decay exponentially. (V = V 0 e k t ) so the terms on the right sides of eqs. (1) and (3) must decay in similar fashion:

I = I 0 e- k t , so di/dt = -k l 0 e" k t and i = y( l - e~kt) so di/dt = ki,e~ k t

By plugging those expressions into eq. (4) and cancelling the exponentials we find that RI 0 - kI 0Lj = kyL 2 . Solving for the unknown constant, k = RI 0/[I 0L, + yL 2 ] = R/[L, + (iyf 0)L 2]. But I 0 = E/(R + r) and if= E/r, so y/I 0 = (R + r)/r]. Therefore, k = R/[L, + L 2(R + r)/r] To check the solution it has to be seen if it is consistent with eq. (2): V 0 e~kt = E - r(I + i) = E - r[I 0e~ k t + y ( l - e - k t ) ] = (E - r i f) + r ( y - I 0 ) e - k t . (E - ry) = 0, so factor out the exponential: V 0 = r ( y - 1 0 ) = r[E/r - E/(R + r)] = E[1 - r/(R + r)] = E[R + r - r]/(R + r) = ER/(R + r). To find the total charge that passes through resistor R, integrate I with respect to time, from t = 0 to 00 Q= Jldt e k t d t IJk = [E/(R + r)][L, + L 2(R + r)/r]/R = (E/R)[L./(R + r) + L 2r] ]

60.

sin 0 = 2R/a = x/(2R) 4x0.1x0.1 x = 4R 2/a = j = 0.04 m = 4 cm ]


hints & solutions mathematics€¦ · ^ bansal classes physics [12] 20 20 (b) wor done by 1k0 v...

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