CBSE Sample Paper for Class 11 Physics Solutions - Set A

AglaSem Schools

SAMPLE QUESTION PAPER – A 1. Practical unit of power is horse power (hp) 1 HP = 746 W 1 2. The stress at which the specimen breaks or ruptures ultimately is called ultimate or tensile strength.1 3. When no exchange of heating energy is possible between the system and surrounding, the process is adiabatic. Such processes are carried in (i) non-conducting cylinders, and (ii) at a fast pace. 1 4. The substance which can be stretched to large values of strain are called elastomers, e.g., elastic tissue of aorta, the largest artery carrying blood from the heart. 5. The necessary and sufficient condition for motion to be simple harmonic is that the restoring for must the linear, i.e., F = – ky or torque, τ = – cθ. 1 6. Joule is a unit of work. Using the relation, work = force × displacement = mass × acceleration × displacement = mass ×

m o c

.

velocity × displacement time

m e s

displacement × displacement = mass × time× time

a l g

= mass × displacement2 × time–2 1 Unit of work, J = kg × m2 × s–2 = kgm2 s–2. 1 7. If a ball is thrown up, the direction of motion of the body is the same as the direction of its velocity whereas the acceleration due to gravity acts on it in the downward direction. Thus, the direction in which an object moves is given by the direction of velocity and not by the direction of acceleration. 1

a . s

l o

o h c

8. Using,

.s

w

and

w

P=

tan α =

P 2 + Q2 +2PQ cos θ Qsin θ P + Q cos θ

(a) R = A + Q and α = 0° (b) R = P – Q and α = 0° to 180° depends on P > Q or Q > P.

w

(c) R =

−1 P 2 + Q 2 and α = tan

½ ½ ½

Q P Or

Component of force along horizontal. Fx = F cos 60° = 72 × Using

Fx = max

We get

ax =

1 = 36 dyne 2

Fx 36 = m 9 = 4 cms–2.

1

1

AglaSem Schools 9. If there is only one propeller, the helicopter that will start rotaing in a direction opposite to that of the rotation of the propeller so as to conserve angular momentum. 2 10. Sound waves are mechanial waves whose velocity γ RT/M

v=

1

Light waves are non-mechanical waves or electromagnetic waves for which c = 1 / m0e 0 , where µ0 is absolute magnetic permeability of free space and ε0 is absolute electrical permittivity of free space. Therefore, v depends upon T, but c does not. 1

mgl3

, g is constant 4 bd3 δ ∴ Maximum relative error in Y is given by

11. Here Y =

∆ Y ∆m 3∆ l ∆ b 3∆ d ∆δ + + + + = ½ Y m l b d δ Thus clearly m, l, b, d and δ introduce the maximum error in the measurement of Y. ½ 12. (a) A lives closer to the school than B because B has to cover higher distance [OP < OQ]. 1 (b) A starts from the school earlier than B because t = 0 for A but B has some finite time. ½ (c) B walks faster than A because it covers more distance in less duration of time [slope of B is grater than that of A]. 1 (d) A and B reach home at the same time. (e) B overtakes A on the road once (at x, i.e., the point of intersection).

m o c

.

m e s

→

a l g

13. The position vector ( r ) of the particle is

→= 3·0t iˆ − 2·0 t 2 ˆj + 4·0 kˆ m r

→

(a) velocity v (t) of the particle is given by

a . s

....(i)

l o

→ → d r d → (r) = v (t ) = dt dt

o h c

.s

w

=

d (3·0t iˆ - 2·0t 2 ˆj + 4·0 kˆ ) dt

= 3iˆ − 4t ˆj + 0

....(ii) 1

w

Also, acceleration → a (t) of the particle is given by ® → d→ d v (t ) v (t ) = a (t) = dt dt d ˆ (3i − 4 t ˆj) = dt = 0 − 4 ˆj

w

→ a (t) = − 4 ˆj

[by using (ii)]

....(iii) ½

(b) At time t, the veocity of the particle is given by using to equation (ii).

→ v (t) = 3iˆ − 4 t ˆj ∴ At t = 2 s,

v = 3·0 iˆ − 4 × 2 ˆj = 3·0 iˆ − 8·0 ˆj

½

AglaSem Schools

∴ Its magnitude is v= =

32 + ( −8)2 =

9 + 64 –1

½

73 = 8·544 ms

and, direction of v is given by,

 vy −8 θ = tan −1   = tan −1     vx  3 = 70° with x-axis. 14. Given, radius of circular bend, r = 30 m Speed of train = v = 54 kmh–1

½

5 18 = 15 ms–1 1 Mass of train, m = 106 kg Anlge of banking = θ = ? The centripetal force is provided by the lateral thrust by the outer rail. According to Newton’s third law of motion, the train exerts (i.e., causes) an equal and opposite thrust on the outer rail causing its wear and tear. ½ The angle of banking by formula, = 54 ×

m o c

.

v2 tan θ = rg

⇒

tan θ =

m e s

½

a l g

(15)2 × 9·8 30

a . s

225 = 0·7653 30 × 9·8 ⇒ tan θ = tan 37° 25′ ∴ = 37°25′ = 37·42°. ½ 15. The ratio of relative velocity of separation after collision to the relative velocity of approach before collision. ½ =

l o

o h c

.s

w

Coefficient of restitution,

w

v2 − v1 e = u −u 1 2

½

where u1 and u2 are initial velocities of the two colliding bodies and v1, v2 are their final velocities after collision. ½ (i) For elastic collision, velocity of separation is equal to the velocity of approach. ∴ e =1 ½ (ii) For inelastic collision, velocity of separation is not zero but always less than the velocity of approach. ∴ 0 θ2. Heat yets conducted in the direction of the fall of temperature. Q∝A ½ Q ∝ (θ1 – θ2) ½ Q∝t C=

Q∝

A(θ1 − θ 2 )t 1 ∝ x x

½

AglaSem Schools

KA (q1 - q 2 )t x

Q=

θ1

½

θ2

x

Here K is a constant called the coefficient of thermal conductivity of the material of the cube and t stands for time interval. We can also write as  ∆θ  H = KA    ∆x  H = Heat flow per second

Where

∆θ = Heat current ∆x T = θ, temperature A = 1 m2 (θ1 – θ2) = 1°C t =1s x = 1 cm θ = K.

m o c

½

m e s

.

a l g

a . s

l o

Then

o h c

½

Or

θ1 − θ 2 dθ or respresents the rate of fall w.r. to d. ½ x dx dθ The quantity represents the rate of change of temperature w.r. to distance and is called dx temperature gradient. The quantity

.s

w

w

w

 dθ  Q = – KA   t  dx 

½

Dimensions of K. Q represents energy and its dimensions are [Q] = [ML2T–2] [dx] = [L] [A] = [L2] [dθ] = [θ] [t] = [T] [K] =

[ML2 T –2 ][L]

[L2 ][q ][T] = [MLT–3 θ–1] 1 (b) Consider a compound wall (or a slab) made of two materials A and B of thickness d1 and d2 . Let K1 and K2 be the coefficients of thermal conductivity of the θ1 and θ2 are the temperature of the end faces (θ1 > θ2) and θ is the temperature of the surface in contact. ½ For material A :

Q1 =

K 1 A1 (θ1 − θ) d1

....(i) 1

For the material B :

AglaSem Schools K 2 A 2 (θ − θ1 ) d2

....(ii) 1

K 1 A 1 (θ1 − θ ) K 2 A 2 (θ − θ1 ) = d1 d2

½

Κ1 θ1 K 2 θ2 + d1 d2 = K1 K 2 + d1 d2

½

Q2 = From eqns. (i) and (ii), we get

Subtracting the value of θ in eqn. (i),

θ=

In general for any number of walls,

θ=

A (θ1 − θ 2 ) d1 d2 + K1 K 2 A θ

B

d1

d2

l o

.s

w

w

w

m e s

a l g

a . s

o h c

.

θ1

A1 (θ1 − θ 2 ) .  d Σ   Κ

m o c ½ ●●