Eduinfy provide you digital online & effective learning for entrance exams .

1 Electrostatics

Two charged particles having charges +25μC and +50μC are separated by a distance of 8 cm. The ratio of forces on them is:

1) 1 : 1

2) 1 : 2

3) 4 : 1

4) 2 : 1

In the given problem, if F is force of repulsion between two charges. Then two charges 25μC and 50μC exerts F force on each other. Hence ratio of forces on them is 1 : 1.

2 Oscillations

The point of suspension of l simple pendulum with normal time period T1 is moving upward according to equation y = Kt^{2} where k = 1m/s^{2}. If new time period is T2 the ratio T_{1}^{2 }/T^{2}_{2} will be :

1) 2/3

2) 5/6

3) 6/5

4) 3/2

SOLUTION

We know that time period of pendulum of l length with net vertical acceleration anet? is given by

Point of suspension moves acoording to y=Kt^{2 }

velocity of pt of susp v=dy? / dt=2Kt

Acceleration of 10 t of susp

3 Work Energy and Power

A worker does 500 J of work on a 10 kg box. If the box transfers 375 J of heat to the floor through the friction between the box and the floor, what is the velocity of the box after the work has been done on it?

1) 5 m/s

2) 10 m/s.

3) 12.5 m/s

4) 50 m/s

Given : $$m = 10$$ kg

Frictional force does negative work i.e $$W_f = -375$$ J

Work done by worker $$W = 500$$ J

Using work-energy theorem : $$W_{all force} = \Delta K.E$$

$$\therefore$$ $$W + W_f = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2$$

OR $$500 - 375 = \dfrac{1}{2}\times 10v^2 - 0$$ $$(\because u = 0 m/s)$$

$$\implies v = \sqrt{25} = 5$$ m/s

4 Laws of Motion

A cart of mass has a block of mass m attached to it as shown in figure. The coefficient of friction between the block and the cart is µ. What is the minimum acceleration of the cart so that the block m does not fall

1)µg

2) g/µ

3)µ/g

4)Mµg/m

**Solution**

Let a be the acceleration of trolley

From the FBD of block

N=ma

∴ µN=mg

µma=mg

a=g/µ

5 Laws of Motion

A cart on an air track is moving at 0.5 m/s when the air is suddenly turned off. The cart comes to rest after traveling 1 m. The experiment is repeated, but now the cart is moving at 1 m/s when the air is turned off. How far does the cart travel before coming to rest?

1) 1 m

2) Ö2 m

3) 2 m

4) 4 m

Solution

1 / 2 mv^{2} = f?x so if velocity is doubled then distance traveled will be four times as great.

6 Ray Optics

In a compound microscope, the focal lengths of two lenses are 2.5 cm and 6.25 cm. An object is placed at 2 cm from objective and the final image is formed at 25 cm from eye lens. The distance between the two lenses is:

1.12 cm

2.15 cm

3.18 cm

4.None of the above

7 Oscillations

Two points P and Q perform SHM along the same straight line, greatest distance between P and Q noticed by an observer studying the motion is found to be A (A=amplitude). Hence, we can deduce that in this case P leads Q ( or Q leads P) by a phase angle of

1) $$\pi$$

2) $$\dfrac { 2\pi }{ 3 }$$

3) $$\dfrac { \pi }{ 2 }$$

4) $$\dfrac { \pi }{ 3 }$$

In one complete oscillation a particle executing S.H.M travels a distance equal to $$4A$$ which is equivalent to $$2\pi$$ phase. In case question the greatest distance between $$P$$ and $$Q$$ is $$A$$. So, they should be separated by a phase$$=\cfrac { A }{ 4A } 2\pi =\cfrac { \pi }{ 2 } $$

Hence $$P$$ leads $$Q$$ by a phase difference of $$\cfrac { \pi }{ 2 } $$

8 Waves

A travelling wave has a velocity of 400 m/s and has a wavelength of 0.5 m. What is the phase difference between two points in the wave that are 1.25 milli secs apart

1) 2π

2) 2π /3

3) 2π/5

4) π/6

The frequency of the wave is f=v/λ=400/0.5=800Hz

Time period = 1/800 = 1.25 ms

Thus, both the points are one time period apart. Hence their phase difference will be zero or 2π

9 Electrostatics

The electroscope used in the electrostatics experiment is typically charged using a power supply with a voltage range:

(1) 0 to 30 volts.

(2) 0 to 100 volts.

(3) 0 to 300 volts.

(4) 0 to 3000 volts..

10 Kinematics

A car travels 90. meters due north in 15 seconds. Then the car turns around and travels 40. meters due south in 5.0 seconds. What is the magnitude of the average velocity of the car during this 20.-second interval?

1) 2.5 m/s

2) 5.0 m/s

3) 6.5 m/s

4) 7.0 m/s

v = d/t.

The DISPLACEMENT of the car is 90 m - 40 m = 50 m, the total time is 20s. V = 50m/20s

NEET

Chemistry

Physics

Biology

Biology

Chemistry

Biology

Chemistry

Biology

Biology

Physics

Physics

Biology

NEET

Chemistry

Biology

Physics

NEET

Biology

NEET

NEET

Biology

Biology

Physics

Chemistry

On

Edu Portal by Planet E tutors for Students and Teachers