MHCET 2013

He must not lost the contact at the highest point, and since he is not losing the contact he must be in the circular motion. 
Equation of motion athe highest point is:
$$N+mg=\dfrac { m{ v }^{ 2 } }{ R } $$
For the minimum velocity the normal has to be zero.
$$\\ mg=\dfrac { m{ { v }^{ 2 } }_{ min } }{ R } \\ \Rightarrow { { v } }_{ min }=\sqrt { Rg } =\sqrt { 5\times 9.8 } =\sqrt { 49 } =7m/s$$
1
#468912
Ask a doubtDoubt
Solution
  1. Report Error
In a circus, a motor cyclist moves in a spherical cage of radius $$5$$m. The minimum velocity with which he must cross the highest point without loosing contact is __________. (Take $$g=9.8m/s^2$$)
A
$$3\ m/s$$
Your Answer
B
$$5\ m/s$$
Your Answer
C
$$6\ m/s$$
Your Answer
D
$$7\ m/s$$
Your Answer
Polarizing andle is defined as the angle of incidence when reflected and refracted ray make an angle of $${ 90 }^{ 0 }$$.
2
#468913
Ask a doubtDoubt
Solution
  1. Report Error
If the light is incident on a transparent medium at a polarising angle then the angle between the reflected and refracted rays is __________.
A
$$30^o$$
Your Answer
B
$$60^o$$
Your Answer
C
$$90^o$$
Your Answer
D
$$180^o$$
Your Answer
We assume that the object in motion just manages to complete the vertical circle. In that case,
$$v_{top} = \sqrt{gR}$$
Total energy at the topmost point $$=\dfrac12mgR + 2mgR = \dfrac52mgR$$
At the horizontal position $$PE = mgR$$
Total energy $$= \dfrac52mgR$$
$$\therefore KE = (\dfrac52-1) = \dfrac32mgR$$
$$\therefore \dfrac{KE}{PE} = \dfrac32$$
3
#468914
Ask a doubtDoubt
Solution
  1. Report Error
In vertical circular motion, the ratio of kinetic energy to potential energy at the horizontal position is ____________.
A
$$5:2$$
Your Answer
B
$$2:1$$
Your Answer
C
$$3:2$$
Your Answer
D
$$2:3$$
Your Answer
Velocity of wave $$=$$ wavelength X frequency
$$v=3\times { 10 }^{ 8 }m/s\\ \lambda =7500\overset { 0 }{ A } =7500\times { 10 }^{ -10 }m\\ f=\dfrac { 3\times { 10 }^{ 8 } }{ 7500\times { 10 }^{ -10 } } =4\times { 10 }^{ 14 }\ Hz$$
4
#468915
Ask a doubtDoubt
Solution
  1. Report Error
If the wavelength of red light in air is $$7500$$ $$\overset{o}{A}$$, then the frequency of light in air is ___________.
A
$$7.5\times 10^{10}$$Hz
Your Answer
B
$$3\times 10^{14}$$Hz
Your Answer
C
$$4\times 10^{14}$$Hz
Your Answer
D
$$5\times 10^{14}$$Hz
Your Answer
Correct equation is:
$$\vec { v } =\vec { \omega  } \times \vec { r } =-\vec { r } \times \vec { \omega  } $$
5
#468916
Ask a doubtDoubt
Solution
  1. Report Error
Which of the following relation is not correct?
A
$$\vec{V}=\vec{\omega}\times\vec{r}$$
Your Answer
B
$$\vec{V}=\vec{r}\times \vec{\omega}$$
Your Answer
C
$$\vec{\delta s}=\vec{\delta \theta}\times \vec{r}$$
Your Answer
D
$$V=r\omega$$
Your Answer
Fringe width is $$\frac { \lambda D }{ d } $$, which is defined as the distance between the two bright fringes. 
Thus, from the center the $$n^{th}$$ bright fring must be at a distance of $$\dfrac { n\lambda D }{ d } $$.
6
#468917
Ask a doubtDoubt
Solution
  1. Report Error
The distance of $$n^{th}$$ bright band on the screen from the central bright band on either sides of central bright band is __________.
A
$$\displaystyle x_n=\frac{n\lambda D}{d}$$
Your Answer
B
$$\displaystyle x_n=(2n-1)\frac{\lambda D}{d}$$
Your Answer
C
$$\displaystyle x_n=\frac{n\lambda d}{D}$$
Your Answer
D
$$\displaystyle x_n=(2n-1)\frac{\lambda D}{2d}$$
Your Answer
Acceleration due to gravity at height $$h$$ is $$g' = \dfrac{gR^2}{(R+h)^2}$$
Let the height be $$h$$ where $$g' =25$$ % of $$g = \dfrac{g}{4}$$
$$\Rightarrow$$ $$\dfrac{g}{4} = \dfrac{gR^2}{(R+h)^2}$$
$$\Rightarrow \dfrac{1}{2}  = \dfrac{R}{(R+h)}$$ 
$$\Rightarrow  h = R$$
7
#468918
Ask a doubtDoubt
Solution
  1. Report Error
The height at which the acceleration due to gravity is $$25\%$$ of that of the surface of earth is ____________. (R=Radius of the earth)
A
$$h=3R$$
Your Answer
B
$$h=2R$$
Your Answer
C
$$h=R$$
Your Answer
D
$$h=\displaystyle\frac{R}{2}$$
Your Answer
$$a\sin\theta =n\lambda \\ \Rightarrow a\sin{ 30 }^{ 0 }=2\lambda \\ a=\dfrac { 2\times 4500\mathring { A }  }{ 1/2 } =18000\mathring { A } $$
8
#468919
Ask a doubtDoubt
Solution
  1. Report Error
Monochromatic light of wavelength $$4500$$ $$\overset{o}{A}$$ falls on slit of width 'a'. In diffraction pattern second maxima deviates through $$30^o$$. The slit width is ________.
A
$$900\overset{o}{A}$$
Your Answer
B
$$18000\overset{o}{A}$$
Your Answer
C
$$13500\overset{o}{A}$$
Your Answer
D
$$22500\overset{o}{A}$$
Your Answer
Geostationary satellites remain at the same position with respect to the Earth, so that they scan the same place in a better way. Therefore, the relative velocity of geostationary satellite with respect to the spinning motion of the Earth is $$0 \ m/s$$.
9
#468920
Ask a doubtDoubt
Solution
  1. Report Error
The relative velocity of geostationary satellite with respect to the spinning motion of the Earth is 
A
$$0 \ m/s$$
Your Answer
B
$$6 \ m/s$$
Your Answer
C
$$12 \ m/s$$
Your Answer
D
$$14 \ m/s$$
Your Answer
Diffraction is defined as the process by which a beam of light or other system of waves is spread out as a result of passing through a narrow aperture or across an edge, typically accompanied by interference between the wave forms produced.
10
#468921
Ask a doubtDoubt
Solution
  1. Report Error
The bending of light near the edges of an obstacle and spreading into the region of geometrical shadow is called ___________.
A
Interference
Your Answer
B
Diffraction
Your Answer
C
Polarization
Your Answer
D
Doppler effect
Your Answer