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#### The product   is equal to :    Option: 1 Option: 2 Option: 3 Option: 4

Sum of an infinite GP

If a is the first term and r is the common ratio of a G.P. Then,

$\mathrm{S_{\infty}=\frac{a}{1-r}}$

$S_{\infty}$ is the sum to infinite terms of the G.P.

Now,

$\\2^{1/4}\cdot4^{1/16}\cdot8^{1/48}\ldots=2^{\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\ldots}\\\Rightarrow 2^{\frac{\frac{1}{4}}{1-\frac{1}{2}}}=\sqrt 2$

#### A particle is moving along the x- axis with its coordinate with time 't' given by $x(t)=10+8t-3t^2$. Another particle is moving along the y-axis with its coordinate as a function of time given by $y(t)=5-8t^3.$ At $t=1$ s, the speed of the second particle measured in the frame of the first particle is given by $\sqrt{v}$. Then $v$ (in m/s) is ____________. Option: 1 580 Option: 2 700 Option: 3 100 Option: 4 300

\begin{aligned} &x_{A}=-3 t^{2}+8 t+10 \\ &\vec{v}_{A}=(-6 t+8) \hat{i} \\ &\text { At time } t=1 \mathrm{~s} \\ &\vec{v}_{A}=2 \hat{i} \\ &Y_{B}=5-8 t^{3} \end{aligned}

$\vec{v}_{B}=-24 t^{2} j$

At time t= 1 s

$\vec{v_{B}}=-24 \hat{j}$

\begin{aligned} &\sqrt{v}=\left|\vec{v}_{B}-\vec{v}_{A}\right|=|-24 \hat{j}-2 \hat{i}| \\ &\sqrt{v}=\sqrt{580} \end{aligned}

So the value of v is 580

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#### The magnifying power of a telescope with tube length $60cm$ is $5$. What is the focal length (in cm) of its eye piece?   Option: 1 30 Option: 2 40 Option: 3 10 Option: 4 20

$\begin{array}{l}{\text { Let the focal length of the objective as } f_{0}} \\ {\text { and focal length of the eyepiece as } f_{e}} \\ {\text { Magnifying power, } M=\frac{f_{0}}{f_{e}}} \\ {\text { Tube length, } L=f_{0}+f_{e}}\end{array}$

$\begin{array}{l}{\text { Given: Magnifying power }=5} \\ \\ {\therefore \frac{f_{0}}{f_{\mathrm{e}}}=5} \\ {\therefore {f}_{0}=5{f}_{e}} \\ and \ {\mathrm{L}=\mathrm{f}_{0}+\mathrm{f}_{\mathrm{e}}=60} \\ {5 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=60\quad\left[\because \mathrm{f}_{0}=5 \mathrm{f}_{\mathrm{e}}\right]} \\ {6\mathrm{f}_{\mathrm{e}}=60} \\ {\mathrm{f}_{\mathrm{e}}=\frac{60}{6}=10 \mathrm{cm}} \\ {\text { So the focal length of eye piece is } 10 \mathrm{cm} .}\end{array}$

So the correct answer is given in option 3.

#### The dimension of stopping potential  in photoelectric effect in units of Planck's constant 'h', speed of light 'c' and Gravitational constant 'G' and ampere A is :    Option: 1   Option: 2  Option: 3    Option: 4

Let

Now,

So,

From here we will get: p-q=1---------(1)

2p+3q+r=2----------(2)

-p-2q-r=-3----------(3)

s=-1-----------(4)

From equation 1, 2,3 and 4 we will get: p=0,q=-1, r=5 and s=-1

So,

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#### The critical angle (in degree)of a medium for a specific wavelength, if the medium has relative permitivity and relative permeability  for this wavelength, will be :  Option: 1 15 Option: 2 30 Option: 3 45 Option: 4 60

and

By using snells law

So the correct option is 2.

#### Consider two solid sphers of radii $R_1=1m,\: R_2=2m$ and masses M1 and M2, respectively. The gravitational field due to sphere  and  are shown. The value of $\frac{M_1}{M_2}$ is:   Option: 1 Option: 2  Option: 3  Option: 4

Considering Case 2 and case 1, one by one

and using $I=\frac{GM}{R^2} \ \ \ \ (\text{Gravitational field on surface of sphere})$

$\dpi{120} \\ 3=\frac{\mathrm{Gm}_{2}}{2^{2}} \\ \\ 2=\frac{\mathrm{Gm}_{1}}{1^{2}} \\ \\ \therefore \frac{3}{2}=\frac{1}{4} \frac{\mathrm{m}_{2}}{\mathrm{m}_{1}} \\ \\ \frac{\mathrm{m}_{1}}{\mathrm{m}_{2}}=\frac{1}{6}$

So the correct option is 2.

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#### The major product $\left [ B \right ]$ in the following sequence of reaction is : $CH_{3}-\underset{CH(CH_{3})_{2}}{C}=CH-CH_{2}CH_{3}\xrightarrow[(i)H_{2}SO_{2},OH^{\ominus })]{(i)B_{2}H_{6}}\left [ A \right ]\xrightarrow[\Delta ]{dil.H_{2}SO_{4}}\left [ B \right ]$ Option: 1     Option: 2     Option: 3     Option: 4

Therefore, Option(3) is correct.

1

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#### Among the reactions (a)  - (d), the reaction(s) that does / do not occur in the blast furnace during the extraction of iron is / are :(a)  $CaO+SiO_{2}\rightarrow CaSiO_{3}$ (b)  $3Fe_{2}O_{3}+CO\rightarrow 2Fe_{3}O_{4}+CO_{2}$ (c)  $FeO+SiO_{2}\rightarrow FeSiO_{3}$ (d)  $FeO\rightarrow Fe+\frac{1}{2}O_{2}$ Option: 1 (C) and (D) Option:2  (A) Option: 3 (A) and (D) Option: 4 (D)

The following reactions do not occur in the blast furnace:

C) $\mathrm{FeO}+\mathrm{SiO}_{2} \rightarrow \mathrm{FeSiO}_{3}$

D) $\mathrm{FeO} \rightarrow \mathrm{Fe}+\frac{1}{2} \mathrm{O}_{2}$

Therefore, Option(1) is correct.

#### The correct order of the calculated spin-only magnetic moments of complexes (A) to (D) is : (A) $\inline Ni(CO)_{4}$ (B) $\inline \left [ Ni(H_{2}O)_{6} \right ]Cl_{2}$ (C) $\inline Na_{2}\left [ Ni(CN)_{4} \right ]$ (D) $\inline PdCl_{2}(PPh_{3})_{2}$ Option: 1      Option: 2  Option: 3  Option: 4

As we have learnt,

$[Ni(CO)_4]$ contains $Ni(0)$ which has a $d^{10}$ configuration having $n=0$ and $\mu=0$

$[Ni(H_2O)_6]^{2+}$contains $Ni^{2+}$ which has a $d^8$ configuration. In a weak ligand field, it has $n=2$ and hence has $\mu=\sqrt8 \ BM$

$[Ni(CN)]_4$contains $Ni^{2+}$which has a $d^8$ configuration. In a strong ligand field, these electrons get paired up and hence $n=0$ and $\mu =0$

$[Pd(PPh_3)_2Cl_2]$ contains $Pd^{2+}$which has a $d^8$ configuration. These electrons get paired in the presence of strong ligand like $PPh_3$ and hence, $n=0$ and correspondingly $\mu=0$.

Therefore, the correct order of spin only magnetic moment of the complexes is

$(B)>(A)=(C)=(D)$

Therefore, Option(4) is correct.