National Physical Laboratory has the responsibility of realising the units of physical measurements based on the International System (SI units) also to realise, establish, maintain, reproduce and update the national standards of measurement & calibration facilities for different parameters.

NiST-F1 ( a cesium fountain clock),

{tex}\eqalign{ & {\text{Here, }}{{\text{R}}_1} = 4 \pm 0.5 \cr & {\text{ }}{{\text{R}}_2} = 12 \pm 0.5 \cr & {\text{Case - I: When }}{{\text{R}}_{{\text{1 }}}}{\text{\& }}{{\text{R}}_{\text{2}}}{\text{ are connected in series:}} \cr & {\text{we know, in series:}} \cr & {{\text{R}}_{net}} = {R_1} + {R_2}{\text{ }} \cr & \Rightarrow {R_{net}} = (4 \pm 0.5) + (12 \pm 0.5) = 16 \pm 1 \cr & \Rightarrow {\text{% Error in }}{{\text{R}}_{net}} = 16 \pm \frac{1}{{16}} \times 100 = 16 \pm 6.25\% \cr & \cr & {\text{Case - II: When }}{{\text{R}}_{{\text{1 }}}}{\text{& }}{{\text{R}}_{\text{2}}}{\text{ are connected in parallel:}} \cr & {{\text{R}}_{{\text{net}}}} = \frac{{{{\text{R}}_{{\text{1 }}}}{{\text{R}}_{\text{2}}}}}{{{{\text{R}}_{{\text{1 }}}} + {{\text{R}}_{\text{2}}}}}{\text{ - (1)}} \cr & {\text{Now,}} \cr & {\text{% Error in }}{{\text{R}}_1} = \frac{{{\text{absolute error in }}{{\text{R}}_1}}}{{{\text{True value of }}{{\text{R}}_1}}} \times 100 \cr & \Rightarrow {\text{% Error in }}{{\text{R}}_1} = \frac{{0.5}}{4} \times 100 = 12.5\% \cr & {\text{Similarly, }} \cr & {\text{% Error in }}{{\text{R}}_2} = \frac{{0.5}}{{12}} \times 100 = 4.16\% \cr & \therefore {\text{ }}{{\text{R}}_1} = 4 \pm 12.5\% \cr & {\text{and }}{R_2} = 12 \pm 4.16\% \cr & \therefore {\text{ }}{{\text{R}}_1}{R_2} = (4{\text{ohm}} \pm 12.5\% )(12{\text{ohm}} \pm 4.16\% ) \cr & {\text{ = (48 oh}}{{\text{m}}^2} \pm 16.66\% ) \cr & and{\text{ }}{R_1} + {R_2}{\text{ }} = 16{\text{ ohm}} \pm 1\% \cr & {\text{Putting these values in equation (1):}} \cr & \Rightarrow {R_{net}} = \frac{{{\text{(48 oh}}{{\text{m}}^2} \pm 16.66\% )}}{{16{\text{ ohm}} \pm 1\% }} = 3{\text{ ohm}} \pm {\text{16}}{\text{.66\% }} \cr} {/tex}

Dimensional formula of wavelength = [M⁰L¹T⁰]

Dimensional formula of frequency = [M⁰L⁰T^-1]

 Intial velocity u = 19.6m/s 

angle = 45

As the football ll move like a projectile, So

Range = {tex}u^2sin2\theta \over g{/tex} = {tex}{u^2sin90^o \over g } = {u^2\over g} {/tex}={tex}{19.6\times 19.6\over 9.8} = 39.2 m{/tex}

Time of Flight = {tex}{2u sin\theta\over g} = {2 \times 19.6 \over 9.8\times \sqrt 2} = {4\over \sqrt 2}{/tex} = {tex}2\sqrt 2{/tex} s

Distacne that the another boy have to cover = 67.4 - 39.2 = 28.2m

Speed of boy = {tex}{28.2\over 2\sqrt 2} = {14.1\over 1.41} = {141\times 100\over 141\times 10} = 10m/s{/tex}

I just hate BERNOULLI

D = 57' = 57' * 60

      = 3240"

arc second to radian = 3240 * 4.85 * 10^-6

                                  = 16587 * 10^-6

D= b/{tex}\theta{/tex} 

D = 6.4 * 10^-6  /  16587 * 10^-6

    = 32/82935

     =   3.85 * 10⁸

Use chain rule for it,

{tex}{d\over dx}(x^3.sin4x){/tex}

{tex}= x^3.{d\over dx}(sin 4x) + sin4x. {d\over dx} (x^3){/tex}

{tex}= x^3.cos4x.{d\over dx}(4x) + sin4x.3x^2{/tex}

{tex}= 4x^3cos4x + 3x^2sin4x{/tex}

Ans. Let g be the value of acceleration due to gravity at the surface of earth and g' at a height h above the surface of earth. If the earth is considered as a sphere of homogeneous composition, then g at any point on the surface of the earth is given by:

{tex}g = {GM\over R^2}{/tex}  …………(1)

Where M mass of earth, R is radius of Earth.

Now, we ll determine Value of g at the distance h from the surface of earth.

Then distance b/w center of earth and point = (R+h)

{tex}g' = {GM\over (R+h)^2}{/tex} ……………(2)

divide eq (2) by (1)

{tex}{g'\over g} ={ { GM\over (R+h)^2} \over {GM\over R^2}}{/tex}

{tex}=> {g'\over g}= {R^2\over (R+h)^2}{/tex}

{tex}=> {g'\over g}= {R^2\over R^2 (1+{h\over R})^2} {/tex}

{tex}=> {g'\over g}= {1\over (1+{h\over R})^2}{/tex}

{tex}=> {g'\over g}= {(1+{h\over R})^{-2}}{/tex}

Expanding using Binomial Theorem and neglecting terms with higher power of R, we get

{tex}=> {g'\over g}=(1-2{h\over R}){/tex}

So, The value of acceleration due to gravity 'g' decreases with altitude 'h'.

When h<< R then expression is

{tex}=> {g'\over g}=(1-2h){/tex}

Ans. {tex}144Km/h = {144\times 1000\over 3600} = 40m/s{/tex}

Ans. The direction of the frictional force will be up the inclined plane in both the motions.

The friction acts opposite to the relative motion of the point of contact on ground. In both the cases the force acting on the point of cylinder in contact with the plane is {tex}mgSin\theta{/tex}.

As there is pure rolling the velocity and acceleration of the point of contact must be zero. So friction acts upwards along the inclined plane.

Ans. Height of tower = 9m

Initial Velocity of drop u = 0

Acceleration Due to gravity = -10m/s 

Let time taken by first drop to reach ground = t

Then, Using

{tex}S = ut +{1\over 2}at^2{/tex}

{tex}=>- 9 = 0\times t + {1\over 2}(-10)t^2{/tex}

{tex}=>- 9 = -5t^2{/tex}

{tex}=> t^2 = {9\over 5} => t = {3\over \sqrt 5}{/tex}

Now, the fourth drop is just leaving the tap, the 2nd and 3rd drop are somewhere midway, and the first drop is hitting the ground. And the time interval for this situation to occur is {tex}{3\over \sqrt 5}{/tex}sec.

Since it says the drops are falling at regular intervals, therefore, the time must be equally divided in that time interval of {tex}{3\over \sqrt 5}{/tex} sec, giving {tex}{1\over \sqrt 5}{/tex} second of intervals between the drops. 

So time taken by second  drop  is {tex} {2\over \sqrt 5} sec{/tex}   and time taken by third drop is {tex}{1\over \sqrt 5}sec{/tex}.

Distance covered by second drop = {tex}{1\over 2} at^2 = {1\over 2}\times 10\times {4\over 5}{/tex}= 4 m

Distance covered by third drop = {tex}{1\over 2} at^2 = {1\over 2}\times 10\times {1\over 5}{/tex}= 1m

So, 2nd drop and 3rd drop are 4m and 1m away from top of tower respectively.

Ans. 

Initial Velocity of balloon u = 0 

Acceleration a = 1.25m/s²

Time t = 10s

Final velocity v = {tex}u + at = 0 + 1.25\times 10 = 12.5m/s{/tex}

Distance traveled in 10 s,

{tex}=> S = ut +{1\over 2}at^2{/tex}

{tex}=> S = 0\times 10 +{1\over 2}\times1.25\times 100{/tex}

=> S = 62.50m 

Now, balloon will drop down with initial velocity, u'= v=12.5 m/s with acceleration, a'= -g =-10 m/s² and cover distance -62.5 m(as it is moving downwards).

Using,

{tex}=> S = ut +{1\over 2}at^2{/tex}

{tex}=> - 62.5 = 12.5t +{1\over 2}(-10)t^2{/tex}

{tex}=> - 62.5 = 12.5t -5t^2{/tex}

{tex}=> 5t^2 - 12.5t - 62.5 = 0{/tex}

{tex}=> t^2 - 2.5t + 12.5 = 0{/tex}

On Solving, We get

t = 5.0 or -2.5

Time can not be negative so the required answer is 5.0 seconds. the stone will reach the ground in 5.0 seconds.

Ans. If an external force is acted on a body to move it, the body accelerates. So the momentum cannot be a constant.

But, if the force cannot make a body to move like push of a boy on a wall, The body remains at rest and the momentum of the body remains zero.

The SI unit of viscosity is Poise.

Please refer this image for your answer.