Bee Quiz

Given, height (h) = 40 m 

Loss of energy = \(\frac{1}{3}rd\) of total mechanical energy

Acceleration due to gravity (g ) = \(10ms^{-2}\)

Letthetotal mechanical energy= 1J

Remaining energy \(=1-\frac{1}{3}=\frac{2}{3}J\)

Now, 

            \(\frac{2}{3}\Big(\frac{1}{2}mv^2+mgh\Big)=mgh\)

\(\Rightarrow \frac{2}{3}\Big(\frac{1}{2}v^2+gh\Big)=gh\)

\(\Rightarrow \frac{v^2}{3}=gh-\frac{2gh}{3}\)

\(\Rightarrow \frac{v^3}{3}=\frac{gh}{3}\Rightarrow v=\sqrt{gh}\)

\(\Rightarrow v=\sqrt{400}\)

\(\Rightarrow v=20m/s\)

Given,

Percentage error in Iength = 3%

Percentage error in force = 4%

We know that,

Pressure, \(P =\frac{F}{A}\)

Area of square \(=(L)^2\)

\(P=\frac{F}{L^2}\)

\(\frac{\Delta P}{P}=\frac{\Delta F}{F}+\frac{2\Delta L}{L}\)

\(\Bigl(\frac{\Delta P}{P}\times 100\Bigr)=\Bigl(\frac{\Delta F}{F}\times 100\Bigr)+2\Bigl(\frac{\Delta L}{L}\times 100\Bigr)\)

\(=4\%+2(3\%)=10\%\)

Total cross-sectional area of the femurs is,

\(A = 2 \times 10\ \text {cm}^2 = 2 \times 10\times 10^{-4}\ \text m^2= 20 \times 10^{-4}\ \text m^2 \)

Force acting on them is

      \( F = mg = 40\ \text {kg} \times 10\ \text { m s}^{-2} = 400\ \text N \)

 \(\therefore \)  Average pressure sustained by them is

       \(P= \frac{F}{A}= \frac{400\ \text N}{20\times 10^{-4}\ \text m^2}=2\times 10^5\ \text N\ \text m^{-2}\)

Function which satisfy the relation \(x^2+y^2=4\) are \(y(x)-\sqrt{4-x^2}\) and \(y(x)=-\sqrt {4-x^2}.\)

And both functions are continuous in \([-2,2].\)

Here, terms greater than  5! i.e.

\((5!)^2,(6!)^2,......,(100!)^2\)  is divisible by 100.

\(\therefore \)  For terms

\((5!)^2, (6!)^2,.....,(100!)^2\)  remainder is 0.

Now, consider \((1!)^2+(2!)^2+(3!)^2+(4!)^2\)

\(=1+4+36+576\)

\(=617\)

When, 617 is divided by 100, its remainder is 17.

So, required remainder is 17.

\(a=\sin^2\theta+\cos^4\theta=\sin^2\theta+(1-\sin^2\theta)^2\)

\(=\sin^4\theta-\sin^2\theta+1+\bigg(\frac{1}{2}\bigg)^2-\bigg(\frac{1}{2}\bigg)^2\)

\(\Rightarrow\ a=\bigg(\sin^2\theta-\frac{1}{2}\bigg)^2+1-\frac{1}{4}=\bigg(\sin^2\theta-\frac{1}{2}\bigg)^2+\frac{3}{4}\)

\(\therefore\ a\geq\frac{3}{4}\)                 \(\bigg[\because\ \bigg(\sin^2\theta-\frac{1}{2}\bigg)^2_{min}=0\bigg]\)

As equatorial magnetic field due to bar magnet is 

\(B_E=\frac{\mu_0m}{4\pi r^3}\)                                               ...(i)

and axial magnetic field due to bar magnet is

\(B_A=\frac{\mu_0}{4\pi}\frac{2}{r^3}\)                                             ...(ii)

From (i) and (ii), we get 

\(B_A=2B_E\)

\(5\times 13=5+13=18,\) then

 \(\frac{18}{2}=(9)^2=81\)

Here, \(\alpha=7x+5,\)

\(\beta=7y+3\)

\(\therefore\ \alpha+\beta=7(x+y)+8=7(x+y)+7+1\)

If \(\alpha+\beta\) is divided by 7, then remainder is 1.

\(\therefore\ r=1\)

Hence, \(\frac{3r+5}{4}=\frac{3+5}{4}=2\)

Let \(\cos^{-1}\Bigl(\frac{1}{2}\Bigr)=\theta \Rightarrow \cos \theta=\frac{1}{2}=\cos\frac{\pi}{3}\)

\(\Rightarrow\ \ \ \theta=\frac{\pi}{3}\in[0,\pi]\)

\(\therefore\)     Principal value of \(\cos^{-1}\Bigl(\frac{1}{2}\Bigr)\) is \(\frac{\pi}{3}\)

Given, \(|a|=8,|b|=3\) and \(|a\times b|=12\)

We know that, \(\sin \theta =\frac{|a\times b|}{|a||b|}\)

\(\Rightarrow \sin \theta =\frac{12}{8\times 3}=\frac{1}{2}\Rightarrow \sin \theta =\sin \frac{\pi}{6}\)

\(\Rightarrow \theta =\frac{\pi}{6}\)                               \(\Big[ \sin \frac{\pi}{6}=\frac{1}{2} \Big]\)

Hence, angle between a and b is  \(\frac{\pi}{6}.\)

Let displacements of masses \(m_1\) and \(m_2\) are \(x_1\) and \(x_2,\) respectively.

Total eloagation of spring is \(x=x_1+x_2.\)

So, when spring snaps back, pull on each of mass is

                 \(F=-kx\)

Hence, by second law equation for \(m_1\) and \(m_2\) are

\(m_1a_1=-kx\Rightarrow m_1 \frac{d^2x_1}{dt^2}=-kx\)

and \(m_2a_2=- kx\Rightarrow m_2 \frac{d^2 x_2}{dt}=-kx\)

Now, from \(x=x_1+x_2,\) we have

     \(\frac{d^2x}{dt^2}= \frac{d^2x_1}{dt_2}+\frac{d^2x_2}{dt_2}\)

\(\Rightarrow \frac{d^2x}{dt_2}=\frac{-k}{m_1}x+\frac{-k}{m_2} x\)

\(\Rightarrow \frac{d^2x}{dt^2}=- k \Big(\frac{1}{m_1}+\frac{1}{m_2}\Big) x\)

So, \(\omega^2=k \Big(\frac{1}{m_1}+\frac{1}{m_2}\Big)= k\Big(\frac{m_1+m_2}{m_1m_2}\Big)\)

Hence, time period of oscillation is

\(T=\frac{2\pi}{\omega}=2\pi\sqrt{ \frac{1}{k}\Big(\frac{m_1m_2}{m_1+m_2}\Big)}\)

Here, \(\phi =10 t^2- 50t+250\)

\(\therefore e=\frac{-d\phi}{dt}=\frac{-d}{dt}(10t^2-50t+250)=-(20 t-50)\)

At \(t=3s,e=-(20\times 3-50)=-10V\)

Here, \(\phi =10 t^2- 50t+250\)

\(\therefore e=\frac{-d\phi}{dt}=\frac{-d}{dt}(10t^2-50t+250)=-(20 t-50)\)

At \(t=3s,e=-(20\times 3-50)=-10V\)

Required no.of operations \(=(2)^{\frac{2(2-1)}{2}}\)

                                      \(=(2)^1=2\)

\(\log _{10}a+\log _{10}b=\log _{10}(a+b)\)

\(\Rightarrow \log _{10}(a.b)=\log _{10}(a+b)\)

\(\Rightarrow a.b=a+b\)

\(\Rightarrow a(b-1)=b\Rightarrow a=\frac{b}{b-1}\)

I astronomical unit \(=1.496\times 10^{11}\text m\)

All the other unit conversions are correct.

Distance moved by the train along circular railway

track\(=\) speed \(\times\) time\(= \Big(\frac{66 \text {km}}{1 \text {hr}}\Big)\times 10 \ \text {sec}\)

\(=\frac{66000\ \text m}{3600 \sec}\times 10\sec=\frac{550}{3}\text m\)

If the angle turned is \(\theta\) radians, then

      \(\theta=\frac{\frac{550}{3}}{1500}=\frac{11}{90}\)

\(\therefore \) Required angle \(=\frac{11}{90}\) radian

                                 \(=\frac{11}{90}\times \frac{180}{\pi}
=\Big(\frac{22}{\pi}\Big)^\circ=7^\circ\)

Distance moved by the train along circular railway

track\(=\) speed \(\times\) time\(= \Big(\frac{66 \text {km}}{1 \text {hr}}\Big)\times 10 \ \text {sec}\)

\(=\frac{66000\ \text m}{3600 \sec}\times 10\sec=\frac{550}{3}\text m\)

If the angle turned is \(\theta\) radians, then

      \(\theta=\frac{\frac{550}{3}}{1500}=\frac{11}{90}\)

\(\therefore \) Required angle \(=\frac{11}{90}\) radian

                                 \(=\frac{11}{90}\times \frac{180}{\pi}
=\Big(\frac{22}{\pi}\Big)^\circ=7^\circ\)

\(y=e^x\) and \(y=\log_ex\) are inverse of each other

Minimum distance of the curve

at (0, 1) and (1, 0)

\(\sqrt{1+1}=\sqrt{2}\)

We have,

\(a=\hat{i}+\hat{j}+\hat{k},\ b=\hat{i}-\hat{j}+\hat{k}\)

\(\Rightarrow\ a\times b=b+\lambda a\)

\(\Rightarrow\ a\cdot(a\times c)=a\cdot b+\lambda|a|^2\)

\(0=1+3\lambda\)

\(\Rightarrow\ \lambda=-\frac{1}{3}\)

\(\Rightarrow\ a\times(b\times c)=a\times b+\lambda(a\times a)\)

\(\Rightarrow\ |a|^2c-(a\cdot c)a=b\times a\)

\(\Rightarrow\ c=\frac{b\times a+(a\cdot c)a}{|a^2|}\)

\(\Rightarrow\ (a\times b)\cdot c=\frac{|a\times b|^2+a\cdot(a\times b)}{|a|^2}\)

\(\Rightarrow\ [a\ b\ c]=\frac{|a\times b|^2}{|a|^2}=\frac{8}{3}\)

We can use newton's law of cooling in convention and mathematically by

\(Q=h A \triangle T\)

Q= rate of conductive heat transfer

h= coefficient of convective heat transfer coefficient

A= area exposed to heat transfer

\(\triangle T=T_s-T_f\) its power is always unity

\(T_s= \) surface temperature

\(T_f=\)fluid temperature

So, the temperature gradient has always unity power

Apply \(C_1 \to xC_1+C_2-C_1\)

=  \(\frac 1 x\)\(\begin{vmatrix}
0 & b & ax+b\\
0& c & bx+c\\
ax+b &bx+c&0
\end{vmatrix}\)

= \(\frac {(ax^2+2bx+c)} {x}\: [b^2x+bc-acx-bc]\)

= \((b^2-ac) (ax^2+2bx+c)\)

= (+ve) (-ve) <0

We have,

\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)

\(\Rightarrow \sqrt{(\sqrt{x-1})^2-2(2)\sqrt{x-1}+(2)^2}\)

     \(-\sqrt{(\sqrt{x-1})^2-2\times 3\sqrt{x-1}+(3)^2}=1\)

\(\Rightarrow \sqrt{(\sqrt{x-1}-2)^2}+\sqrt{(\sqrt{x-1}-3)^2}=1\)

\(\Rightarrow |\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1\)

                                                   \(x\in [5,10]\)

    \(\sqrt{x-1}-2-\sqrt{x-1}+3=1\)

\(\therefore 1=1\)

Hence, x has infinite solutions in  \(x\in [5,10]\).

Given,

\(2\sin\alpha + 3\cos\beta =3\sqrt 2\)                     ...(i)

\(3\sin\beta+2\cos\alpha=1\)                          ...(ii) 

On squaring and adding Eqs. (i) and (ii), we get

\(4+ 9+ 12(\sin\alpha\cos\beta+ \sin\beta\cos\alpha)=18+ 1\)

\(\Rightarrow 12\sin(\alpha + \beta) = 6\)

\(\Rightarrow \sin(\alpha+\beta)=\frac{1}{2}\)

\(\Rightarrow \alpha+\beta=150^\circ\)

\(\Rightarrow \alpha+\beta\ne 30^\circ\)

\(\therefore \gamma=180-(\alpha+\beta)\)

\(\Rightarrow \gamma=30^\circ \)

The wave number (\(\bar v\)) of the radiation

\(=\frac{1}{\lambda}\)

\(=R_\infty \Bigl[\frac{1}{n^2_1}-\frac{1}{n^2_2}\Bigr]\)

Now for case \((\text I) \ n_1= 3, n_2 = 2 \)

\(\frac{1}{\lambda_1}=R_\infty\Bigl[\frac{1}{9}\frac{-1}{4}\Bigr],R_\infty=\text{ Rydberg constant } \)

\(\frac{1}{\lambda_1}=R_\infty\Bigl[\frac{4-9}{36}\Bigr]=\frac{-5R_\infty}{36}\)

\(\Rightarrow \lambda_1=\frac{-36}{5R_\infty} \)

\(\frac{1}{\lambda_2}=R_\infty\Bigl[\frac{1}{4}-\frac{1}{1}\Bigr]=\frac{-3R_\infty}{4}\Rightarrow \lambda_2=\frac{-4}{3R_\infty}\)

\(\Rightarrow \frac{\lambda_1}{\lambda_2}=\frac{-36}{5R_\infty}\times \frac{3R_\infty}{-4}\Rightarrow \frac{\lambda_1}{\lambda_2}=\frac{27}{5}\)

Area under F-f curve will give change in momentum

\(\therefore\) According to the graph in question = Area = 12 Ns.

I. \(\lim\limits_{n\to\infty}\frac{2^n+(-2)^n}{2^n}=\lim\limits_{n\to\infty}1+(-1)^n=\) does not exist

II. \(\lim\limits_{n\to\infty}\frac{3^n+(-3)^n}{4^n}=\lim\limits_{n\to\infty}\Bigl(\frac{3}{4}\Bigr)^n+\Bigl(\frac{-3}{4}\Bigr)^n\)

\(=0+0=0\)

When the magnet is pushed into the coil, magnetic flux linked with the coil changes. An emf is induced in the coil, which produces maximum deflection.