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The product  2^{1/4}\cdot 4^{1/16}\cdot 8^{1/48}\cdot 16^{1/128}\cdot \cdots is equal to :   
Option: 1 2^{1/4}
Option: 2 2
Option: 3 2^{1/2}
Option: 4 1
 

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

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Posted by

avinash.dongre

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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Posted by

vishal kumar

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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.

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Posted by

vishal kumar

The dimension of stopping potential V_0 in photoelectric effect in units of Planck's constant 'h', speed of light 'c' and Gravitational constant 'G' and ampere A is :   
Option: 1 h^{-2/3}c^{-1/3}G^{4/3}A^{-1}
 
Option: 2 [h]^0[G]^{-1}[c]^5[A]^{-1}

Option: 3 h^{2}G^{3/2}c^{1/3}A^{-1}  

Option: 4 h^{1/3}G^{2/3}c^{1/3}A^{-1}
 

 

 

 Let V_0=[h]^p[G]^q[c]^r[A]^s

Now, K_{max}=eV_0\Rightarrow [K_{max}]=[eV_0]\Rightarrow [V_0]=\frac{[K_{max}]}{[e]}\Rightarrow [V_0]=\frac{[ML^2T^{-2}]}{[AT]}\therefore [V_0]=[ML^2T^{-3}A^{-1}]

E=hf\Rightarrow [h]=\frac{[E]}{[f]}\Rightarrow \frac{[ML^2T^{-2}]}{[T^{-1}]}\Rightarrow [h]=[ML^2T^{-1}]

 

F=\frac{Gm^2}{r^2}\Rightarrow [G]=\frac{[F][r^2]}{[m^2]}=\frac{[MLT^{-2}][L^2]}{[M^2]}=[M^{-1}L^3T^{-2}]

So, [ML^2T^{-3}A^{-1}]=[ML^2T^{-1}]^p[M^{-1}L^3T^{-2}]^q[LT^{-1}]^r[A]^s

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, [V_0]=[h]^0[G]^{-1}[c]^5[A]^{-1}

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Posted by

vishal kumar

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

 

 

 

 

 

C_{vacuum}=\frac{1}{\sqrt{\mu _0\epsilon _0}} \ and \ C_{medium}=\frac{1}{\sqrt{\mu _0\mu _r\epsilon _0\epsilon _r}}

and  n=\frac{C_{vacuum}}{C_{medium}}=\sqrt{\mu _r\epsilon _r} =\sqrt{\frac{4}{3}*3}=2

By using snells law 

n*sinC=1*sin(90^0)\\ \Rightarrow n*sinC=1\\ \Rightarrow sinC=\frac{1}{2}\Rightarrow critical \ angle =C=30^0

So the correct option is 2.

 

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Posted by

vishal kumar

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 \frac{2}{3}

Option: 2 \frac{1}{6}

Option: 3 \frac{1}{2}

Option: 4 \frac{1}{3}
 

Considering Case 2 and case 1, one by one

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

R-Radius of sphere

                                                            \\ 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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Posted by

vishal kumar

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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.

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Posted by

Kuldeep Maurya

Which of the following compounds is likely to show both Frenkel abnd Schottky defects in its crystalline form ?
Option: 1 AgBr

Option: 2 CsCl

Option: 3 KBr

Option: 4 ZnS
 

1

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Posted by

Aarchi Jain

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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.

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Posted by

vishal kumar

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

Option: 3 (C)< (D)< (B)< (A)

Option: 4 (A)\approx (C)\approx (D)< (B)
 

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)]_4contains 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.

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Posted by

vishal kumar

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