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1 Gravitation

The height at which the weight of a body becomes 1/16 th its weight on the surface of (radius R) is

1) 4R

2) 3R

3) 15R

4) 5R

**SOLUTION**

2 Laws of Motion

An elevator weighing 6000 kg is pulled upward by a cable with an acceleration of 5ms^{-2} . Taking g to be 10ms-2 , then the tension in the cable is

1) 6000 N 2) 9000 N 3) 60000 N 4) 90000 N

Solution :

T = m (g + a) = 6000 (10 + 5) T = 90,000 N

3 Waves

The Equation of a wave on a string of linear mass density 0.04 Kg m^{-1 } is given by

1) 4.0 N

2) 12.5 N

3) 0.5 N

4) 6.25 N

**SOLUTION**

4 Gravitation

Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero, is

SOLUTION

5 Mechanical Properties Of Fluids

Tanks A and B open at the top contain two different liquids upto certain height in them. A hole is made on the wall of each tank at a depth 'h' from the surface of the liquid. The area of the hole in 'A' is twice that of in B. If the liquid mass flux through each hole is equal, then the ratio of the densities of the liquids respectively is

**SOLUTION**

6 Gravitation

An object weights 72 N on the earth. Its weight at a height R/2 from earth is = .............. N

1) 72

2) 0

3) 56

4) 32

**SOLUTION**

7 Ray Optics

Let the x - z plane be the boundary between two transparent media. Medium 1 in Z ≥ 0 has a refractive index of Ö2 and medium 2 with z < 0 has a refractive index of Ö3. A ray of light in medium 1 given by the vector is incident on the plane of separation. The angle of refraction in medium 2 is

1) 45°

2) 60°

3) 75°

4) 30°

**SOLUTION**

The field created by the magnets alone at the point O

8 Ray Optics

A double convex lens made of glass of refractive index $$1.56$$ has both radii of curvature of magnitude $$20 \,cm$$. If an object is placed at a distance of $$10 \,cm$$ from this lens, the position of the image formed is

1) $$22.86$$ same side of the object

2) $$22.86$$ opposite side of the object

3) $$44.89$$ same side of the object

4) $$44.89$$ opposite side of the object

Solution:

Here, $$R_1 = 20 \,cm, R_2 = - 20 \,cm, u = -10\, cm$$

and $$\mu = 1.56$$

Using lens make'rs formula,

$$\frac{1}{f} = \left(1.56 -1\right)\left(\frac{1}{20}+\frac{1}{20}\right)$$

$$ \Rightarrow f = \frac{20}{0.56\times2} = 17.86 \,cm $$

Now, from lens equation,

$$v = \frac{uf}{u+f} = \frac{-10\times 17.86}{-10+17.86} = -22.86 \,cm$$

Since $$v$$ is negative, the image will be formed on the same side as that of object.

9 Thermal Properties Of Matter

A lead bullet strikes against a steel plate with a velocity 200 m/s. If the impact is perfectly inelastic and the heat produced is equally shared between the bullet and the target, then the rise in temperature of the bullet is (specific heat capacity of lead =125 Jkg ^{-1} K ^{-1})

1) 80° C

2) 60° C

3) 40° C

4) 120° C

**SOLUTION**

10 Kinematics

A particle moves along positive branch of the curve y=x/ 2 where x=t3/ 3, x and y are measured in metres and t in seconds, then

1) The acceleration of particle at t=2s is iˆ+2jˆ

2) The acceleration of particle at t=1s is 2iˆ+ jˆ

3) The velocity of particle at t=2s is iˆ+1/2 jˆ

4) The velocity of particle at t=1s is 1/2 iˆ+ jˆ

SOLUTION

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