Physics - Paper I - For Class XI (Science Group) - Model papers With Solved MCQs 2020 -2021

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Physics - Paper I
For Class XI (Science Group)
Model papers 2020 -2021

Max. Marks:43 (25 Marks)

SECTION B (SHORT-ANSWER QUESTION)

NOTE: Attempt any Five part questions from this section. All questions carry equal marks. The use of scientific calculator is allowed. All notations are used in their usual meanings. Draw diagram where necessary.
Q.2 (i): State and prove the law of Conservation of Linear Momentum.
Ans: Law of Conservation of Linear Momentum
Introduction:
This law deals with the momentum of the bodies can collide with each other.

Statement:
"When some bodies in an isolated system acts upon one another, the total momentum of the system remains constant."

OR

"The total momentum of an isolated system of interacting bodies remain constant."

Explanation:
According to this law if two or more bodies in an isolated system collide with one another the total momentum of the system before collision is equal to the total momentum of the system after collision.

Mathematical representation:
Two bodies of masses m₁ and m₂ moving with velocities u₁ and u₂ collide with each other and their velocities becomes v₁ and v₂, then according to the law of momentum.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Proof:
Consider two bodies a and b of masses m₁ and m₂ moving with velocities u₁ and u₂ in same direction as shown in figure.
Let the bodies collide for "t" second and experience force "f". Due to which their velocities become v₁ and v₂.
During collision body a exerts some force f on the body b.The force exerted by body a on body b is equal to the rate of change of momentum of body a i.e.,
Force on body a= rate of change of momentum of body a

m₁v₁ - m₁u₁ = - m₂v₂ + m₂u₂

m₁v₁ + m₂v₂ = m₁u₁ + m₂u₂

(ii) What is difference between static and dynamic equilibrium? State the conditions of equilibrium.
Ans: Difference Between Static And Dynamic Equilibrium
Static Equilibrium:
If the combined effect of all the forces acting on a body is zero and the body is in the state of rest then its equilibrium is termed as static equilibrium.
For example,

• A book lying on a table
• A block hung from a string

Dynamic Equilibrium:
The equilibrium of bodies moving with uniform velocity is called dynamic equilibrium.
For example,

• The jumping of a paratrooper by a parachute is an example of uniform motion, In this case weight is balanced by the reaction of the air on the parachute acts in the vertically upward direction.
• The motion of a small steel ball through a viscous liquid. Initially the ball has acceleration but after covering a certain distance, its velocity becomes uniform because weight of the ball is balanced by upward thrust and viscous force of the liquid. Therefore, ball is in dynamic equilibrium.

Conditions of Equilibrium

There are two conditions of equilibrium:
First Condition of Equilibrium:
According to 1^st condition of equilibrium:-
"An object is said to be in equilibrium if the sum of all the force acting on it in one direction balances the sum of all forces in opposite direction."

Or

"The resultant of the forces acting on the body is zero."

Mathematical Representation:-
If an object is acted upon by forces along x-axis and y-axis then according to 1^st condition of equilibrium:
The sum of all the force acting along x-axis is 0.

∑ F_x = 0
Or  F_1x + F_2x + F_3x .................... F_10x = 0

Similarly, The sum of all the force acting along y-axis is 0 i.e.

∑ F_y = 0

Or  F_1y + F_2y + F_3y .................... F_10y = 0

Since forces along x-axis and y-axis are 0 therefore total force will be zero.

∑ F = 0

Second Condition of Equilibrium:-
Second condition of equilibrium deals with such object where torque can produce.
According to 2_nd condition of equilibrium:-
"If sum of all the torque acting on a body is zero then body is in equilibrium."

Or

"A body is in equilibrium when sum of all the clockwise torque is equal to sum of all the anticlockwise torque."

Mathematical Representation:-

∑ 𝜏 = 0

Clockwise Torque = Anticlockwise Torque

(iii) Drive an expression for the Variation of "g" with depth.
Ans:

(iv) How is the magnifying power of the (i) Astronomical telescope and (ii) compound microscope affected by increasing the focal length of their objectives?
Ans: (i) Astronomical Telescope:
The magnifying power of the Astronomical telescope is given by:

M = f_o / f_e

where
f_o = focal length of objective
f_e = focal length of eyepiece
If the focal length of eyepiece remains the same, then:

M ∝ f_o

Which shows that magnifying power of the Astronomical telescope is directly proportional to the focal length of their objectives. It will increase if the focal length of their objectives increase and the magnifying power of Astronomical telescope will decrease if the focal length of their objectives decrease.

(ii) Compound Microscope:
The magnifying power of the compound microscope is given by:

M = ^L/ _fo (1 + ^d/_f)

where
f_o = focal length of objective
f_e = focal length of eyepiece
d = least distance of distant vision
If the focal length of eyepiece and least distance remains constant, then:

M ∝ ¹ / _fo

Which shows that magnifying power of the compound microscope is inversely proportional to the focal length of their objectives. It will increase if the focal length of their objective decreases and the magnifying power of compound microscope will decreases if the focal length of their objective increases.

(v) Prove that the vectors -------------- can form the sides of a right angled triangle.
Solution:

(vi) Two coherent sources are placed 1.8cm apart. Interference fringes are obtained on screen 80cm away. The fourth bright fringe is at a distance of 1.08 cm from the central fringe. Calculate the wavelength of the light used.
Data:
d = 1.8 cm
L = 80 cm
n = 4
y₄ = 1.08 cm
λ = ?

Solution:
The position of bright fringe with central fringe is given by:
y₄ = ^{n λ L} /_d
1.08 = ^{(4) λ (80)} /_1.8
1.08 x 1.8 = 320 λ
λ = 1.08 x 1.8 / 320
λ = 0. 006075 cm = 0. 006075 / 100
λ = 6.075 x 10^-5 m Ans.

(vii) Find the speed of sound in air at 50 °C and 70 °C (take speed of sound 332 m/s).
Data:
T₁ = 50 °C
T₂ = 70 °C
V_o = 332 m/s
V₅₀ = ?
V₇₀ = ?

Solution:
The speed of sound at 1 °C
V_t = V_o + 0.61t

At 50 °C:
V₅₀ = 332 + (0.61 x 50)
V₅₀ = 362.5 m/s Ans.

At 70 °C:
V₅₀ = 332 + (0.61 x 70)
V₅₀ = 374.7 m/s Ans.

(viii) A truck starts from rest at the top of a slope which is 1 m high and 49 m long. Find its acceleration and speed at the bottom of the slope assuming that friction is negligible. ( Chapter 3: Motion -  example 3.7 - from text book)
Data:

• V_i = 0
• h = 1 m
• l = 49 m
• a = ?
• V_f = ?

Solution:
Since there is no motion perpendicular to the plane for R and Wcos θ must balanced each other as shown in figure:

∴ R - Wcos θ = 0

When friction is negligible, The only unbalanced force acting on the truck is Wsinθ  acting along the X-axis or parallel to the plane. If produces an acceleration (a) and (m) be the mass of the truck, then by Newton's 2nd Law of motion.

W sin θ = ma
But W = mg
∴ mg sin θ = ma
or a = g sin θ
Sin θ = ^h / _l = ¹ / ₄₉
(as sin θ = ^{perpendicular} / _hypotenuse = ^h / _l)
∴ a = 9.8 x ¹ / ₄₉ = 0.2 m/s²

To determine the speed at the bottom of the slope we make use of the equation of motion. We have,

• V_i = 0
• a = 0.2 m/s²
• S = 49 m
• V_f = ?

∴ V_f² - V_i² = 2as
V_f² = V_i² + 2as
V_f² = (0)² + 2 x 0.2 x 49
V_f² = 19.6
Taking square root on both side
∴ V_f = √ 19.6
V_f = 4.4 ms^-1 Ans.

(ix) A diver leaps from a tower with an initial horizontal velocity component of 7 m/s and upward velocity component of 3 m/s. Find the component of her position after 1 second.
Data:
V_ox = 7 m/s
V_oy = 3 m/s
t = 1sec
x-component = ?
y-component = ?

Solution:
For x-component:
S = V x t
x = V_ox x t
x = 7 x 1
x = 7 m Ans.

For y-component:
S = V_it + ¹/₂ at²
y = V_oyt + ¹/₂ (-g)t²
y = 3 x 1 - ¹/₂ x 9.8 (1)²
y = 3 - 4.9
y = - 1.9 m Ans.
The value of y-component is negative, which show downward direction.



SOURCE: Board Of Intermediate Education Karachi