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# Tutor profile: Sehaj N.

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Sehaj N.
Certified Tutor with 10 yrs of TEACHING EXPERIENCE
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## Questions

### Subject:Physics (Waves and Optics)

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Question:

During the Young’s Double Slit Experiment (YDSE), suppose a glass slab of thickness \$\$“t”\$\$ & refractive index \$\$“µ”\$\$ is placed in front of one of the slits. (a) Explain the change in the Interference pattern observed. (b) Suppose the first dark fringe is formed at the center, find the value of the wavelength of the light used.

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Sehaj N.

(a) Let, the velocity of light in the air = \$\$“v”\$\$ When a glass slab of thickness \$\$“t”\$\$ is placed in the path of light its velocity will decrease by \$\$“µ”\$\$ times. For the time interval when light from Slit 1 passes through-thickness \$\$“t”\$\$, its velocity will be \$\$“v/µ”\$\$. It will travel a distance \$\$“t”\$\$ For the same time interval when light from Slit 2 passes through the air, its velocity will be \$\$“v”\$\$. It will travel a distance \$\$“tµ”\$\$ (here we have simply compared using formula \$\$distance = speed×time\$\$) Therefore, when light from both the slits reaches the center they will have a phase difference of : \$\$tµ – t = t(µ – 1)\$\$ The interference pattern will shift by \$\$t(µ – 1)\$\$ (b) For a dark fringe to form both lights reaching from slit 1 and slit 2 should interfere destructively. So the phase difference between them should be \$\$180^O\$\$ or \$\$λ/2\$\$ (where \$\$λ\$\$ is the wavelength of light). Therefore : \$\$ t(µ – 1) = λ/2\$\$ \$\$λ = 2t(µ – 1)\$\$

### Subject:Physics (Newtonian Mechanics)

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Question:

We have taken a heavy stone and light feather to the surface of the moon & dropped them simultaneously from the same height. What will reach the ground first: Stone or Feather? If the same experiment is conducted on the surface of Earth will there be any change in observation and why?

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Sehaj N.

ON THE SURFACE OF THE MOON The only force acting on both the object is Gravitational force According to Newton’s law of Gravitation, Force acting on both objects will be: \$\$(G × Mass of Stone or Feather × Mass of Moon)/r^2 = Mass of Stone or Feather × Acceleration\$\$ Therefore in both cases Acceleration comes out to be \$\$(G ×Mass of Moon)/r^2\$\$ which is constant. (r can approx. be taken to be the radius of Moon) \$\$a=g_m\$\$ (acceleration due to gravity on moon) For both cases, Stone & Feather, the value of acceleration is the same. Now both Stone & Feather are dropped (\$\$ u = 0 \$\$) from the same height (s is same): \$\$s=ut+1/2 at^2\$\$(all variables \$\$"s","u" and "a"\$\$ are same for both cases) From here we can conclude that “t” will be the same for both cases. Therefore, Both Stone & Feather will reach the ground AT THE SAME TIME. ON THE SURFACE OF THE EARTH We can follow steps as previously done BUT THERE IS ONE MORE FORCE ACTING. The atmosphere of the moon is a vacuum. I.e. no air is present. But the atmosphere of the Earth has “air” on it. When Stone or Feather are dropped, alongside the gravitational force, air friction will also act. \$\$(G × Mass of Stone or Feather × Mass of Moon)/r^2 – Frictional Force = Mass of Stone or Feather ×Acceleration\$\$ Here, in this case, the value of acceleration in both cases will be different. If we assume that Frictional force is the same for both Feather & Stone, then \$\$a=g_e- F/M\$\$ (\$\$g_e=\$\$ acceleration due to gravity on earth) Now in the case of Stone M is more; Therefore F/M is less; Hence "\$\$a\$\$" will be more. acceleration for stone > acceleration for feather Both stone and feather are dropped (\$\$u = 0\$\$) from the same height (s is the same) \$\$s=ut+1/2 at^2\$\$ For the case of stone "\$\$a\$\$" is more (\$\$ t= √(2s/a\$\$) ); t will be less. We can conclude STONE WILL REACH THE EARTH THE FIRST IF WE TAKE INTO ACCOUNT THE FRICTIONAL FORCE BECAUSE OF AIR

### Subject:Calculus

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Question:

For the equation, \$\$y= x^2-6x+10\$\$, find the minimum value of y possible.

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Sehaj N.

This question can be solved by two methods: Method 1: \$\$y= x^2-6x+10\$\$ Differentiating y with respect to x will give us the slope of y \$\$dy/dx =2x-6\$\$ We can note that \$\$y= x^2-6x+10\$\$ Is a quadratic equation having two roots possible. The slope represents the change in the value of y with respect to x. \$\$dy/dx =2x-6=0\$\$ \$\$2x-6=0\$\$ \$\$2x=6\$\$ \$\$x=3\$\$ Now we can put any value of x before 3 and after 3 to note the value of the slope. Let \$\$x=2\$\$ then \$\$dy/dx=2(2)-6=4-6= -2\$\$ ( a negative value) Let \$\$x=4\$\$ then \$\$dy/dx=2(4)-6=8-6= 2\$\$ ( a positive value) Now we note : \$\$dy/dx >0\$\$ for \$\$x >3\$\$ \$\$dy/dx=0\$\$ for \$\$x=3\$\$ \$\$dy/dx <0\$\$ for \$\$x <3\$\$ We can note y will decrese till \$\$x = 3\$\$, then it will start increasing Therefore at \$\$x = 3\$\$; the value of y is minimum \$\$y= x^2-6x+10\$\$ \$\$y= (3)^2-6(3)+10\$\$ \$\$y= 9-18+10\$\$ \$\$y= 1\$\$ It is the minimum value of y possible. Method 2 \$\$y= x^2-6x+10\$\$ \$\$dy/dx =2x-6=0\$\$ \$\$2x-6=0\$\$ \$\$2x=6\$\$ \$\$x=3\$\$ \$\$dy/dx =2x-6\$\$ \$\$d^2 y/dx^2 =2\$\$ ( Positive ) (\$\$ d^2 y/dx^2 >0\$\$ Condition for Minimima ; \$\$d^2 y/dx^2 <0\$\$ Condition for Maxima ) Therefore at \$\$x = 3\$\$; the value of y is minimum \$\$y= x^2-6x+10\$\$ \$\$y= (3)^2-6(3)+10\$\$ \$\$y= 9-18+10\$\$ \$\$y= 1\$\$ It is the minimum value of y possible.

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