CBSE 11TH PHYSICS - Online Test
Q1.
In constructing a large mobile, an artist hangs an aluminum sphere of mass 6.0 kg from a vertical steel wire 0.50 m long and 2.5 in cross-sectional area. On the bottom of the sphere he attaches a similar steel wire, from which he hangs a brass cube of mass 10.0 kg. Compute the tensile strain.
Answer : Option A
Explaination / Solution:
No Explaination.
Q2. A washer consists of a 3.00 cm diameter circle of sheet metal with a 1.00 cm diameter circular hole in the middle. If the metal washer is heated until the diameter of the washer is 3.03 cm, then the diameter of the hole will be
Answer : Option D
Explaination / Solution:
diameter of hole = 1+0.01 = 1.01cm
Q3. Approximately How many light years is a meter?
Answer : Option D
Explaination / Solution:
A light-year is a unit of distance. It is the distance that light can travel in one year. Light moves at a velocity of about 300,000 kilometers (km) each second. So in one year, it can travel about 10 trillion km. More p recisely, one light-year is equal to
=> 1 Light Year =
=> 1 m =
Q4. A spring has a certain mass suspended from it and its period for vertical oscillations is . The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now .The ratio of / is
Answer : Option D
Explaination / Solution:
No Explaination.
Q5. A truck is coming towards you from a distance of 1000m on a straight road at time t=0. At time t=10 seconds it is at a distance of 900 m. The average velocity in m/sec is
Answer : Option C
Explaination / Solution:
Initial distance x1 = 1000m
Initial time t1= 0 sec
Final distance x2 = 900 m
Final time t2 = 10 sec
Average velocity
=
Q6. Velocity of sound in air is 300 m/s. Then the distance between two successive nodes of a stationary wave of frequency 1000 Hz is.
Answer : Option A
Explaination / Solution:
No Explaination.
Q7. A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?
Answer : Option D
Explaination / Solution:
Maximum horizontal distance, R = 100 m
The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°, i.e., θ = 45°.
The horizontal range for a projection velocity v, is given by the relation:
R =( u2 Sin 2θ / g)
100 =
=> = 100 ….(i)
The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H.
Acceleration, a = –g
Using the third equation of motion:
v2 – u2 = -2gH
The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°, i.e., θ = 45°.
The horizontal range for a projection velocity v, is given by the relation:
R =( u2 Sin 2θ / g)
100 =
=> = 100 ….(i)
The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H.
Acceleration, a = –g
Using the third equation of motion:
v2 – u2 = -2gH
Q8. A steel rod 2.0 m long has a cross-sectional area of 0.30 . It is hung by one end from a support, and a 550-kg milling machine is hung from its other end. Determine the elongation. Take Young's modulus of steel as 20 Pa
Answer : Option C
Explaination / Solution:
No Explaination.
Q9. Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. What is the change in internal energy of the gas?
Answer : Option B
Explaination / Solution:
Internal energy depends on temperature and temperature remains constant because system is thermally insulated. so that change in internal energy will be zero.
Q10. A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s-2. Calculate the initial thrust (force) of the blast.
Answer : Option A
Explaination / Solution:
Initial thrust = upthrust required to impart acceleration + uthrust to overcome gravity
=
