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A massless string connects two pulley of masses ' 2 \mathrm{~kg}' and '1 \mathrm{~kg}' respectively as shown in the figure.

The heavier pulley is fixed and free to rotate about its central axis while the other is free to rotate as well as translate. Find the acceleration of the lower pulley if the system was released from the rest. [Given, g=10 \mathrm{~m} / \mathrm{s}^2]

Option: 1

\frac{4}{3} \mathrm{~gm} / \mathrm{s}^2


Option: 2

\frac{3}{2} \mathrm{~gm} / \mathrm{s}^2


Option: 3

\frac{3}{4} \mathrm{~gm} / \mathrm{s}^2


Option: 4

\frac{2}{3} \mathrm{~gm} / \mathrm{s}^2


3/4gm/s2

 

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Posted by

Guru G

Calculate the acceleration of block m_1 of the following diagram. Assume all surfaces are frictionless . Here m1 = 100kg and m2 = 50kg

 

Option: 1

0.33m/s2


Option: 2

0.66m/s2


Option: 3

1m/s2


Option: 4

1.32m/s2


0.66m/s2

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Posted by

Guru G

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No. of transition state in given figure

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


2

 

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Posted by

Mura

When cell has stalled DNA replication fork, which checkpoint should be predominantly activated?

Option: 1

G1/S

 

Option: 2

G2/M

 

 

 

Option: 3

M

 

 

Option: 4

Both G2 M and M

 

G2/M should be activated as the cell has stalled DNA replication fork.

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Posted by

Ajit Kumar Dubey

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A 100 \; m long wire having cross-sectional area 6.25 \times 10^{-4}m^{2} and Young's modulus is 10^{10}Nm^{-2}  subjected to a load of 250\; N, then the elongation in the wire will be:
Option: 1 4 \times 10^{-3} \mathrm{~m}
Option: 2 6.25 \times 10^{-3} \mathrm{~m}
Option: 3 6.25 \times 10^{-6} \mathrm{~m}
Option: 4 4 \times 10^{-4} \mathrm{~m}

\begin{aligned} & \text { Stress }=\mathrm{y} \text { strain } \Rightarrow \frac{W}{\mathrm{~A}}=\mathrm{y} \frac{\Delta \ell}{\ell} \\ & \Delta \ell=\frac{\mathrm{W} \ell}{\mathrm{yA}} \Rightarrow \Delta \ell=\frac{250 \times 100}{10^{10} \times 6.25 \times 10^{-4}} \\ & \Delta \ell=4 \times 10^{-3} \mathrm{~m} \end{aligned}

Hence, the correct answer is option 1

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Posted by

rishi.raj

A circular loop of radius r is carrying current I\; A. The ratio of the magnetic field at the center of circular loop and at a distance r from the center of the loop on its axis is:
Option: 1 2 \sqrt{2}: 1
Option: 2 1: 3 \sqrt{2}
Option: 3 1: \sqrt{2}
Option: 4 3 \sqrt{2}: 2

Magnetic field at centre of coil B_1=\frac{\mu_0 I}{2 r}

on the axis at

 x=r \Rightarrow B_2=\frac{\mu_0 \mathrm{Ir}^2}{2\left(r^2+x^2\right)^{3 / 2}}

\begin{aligned} & \mathrm{B}_2=\frac{\mu_0 \mathrm{Ir}^2}{2\left(\mathrm{r}^2+\mathrm{r}^2\right)^{3 / 2}} \\ & \mathrm{~B}_2=\frac{\mu_0 \mathrm{I}}{2(2 \sqrt{2} r)} \\ & \frac{\mathrm{B}_1}{\mathrm{~B}_2}=2 \sqrt{2} \end{aligned}

\(2 \sqrt{2}: 1\)

 

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Posted by

rishi.raj

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In the expansion of \left ( \frac{x}{cos\theta }+\frac{1}{x\sin \theta } \right )^{16}, if  L_{1}  is the least value of the term independent of x when \frac{\pi }{8}\leq \theta \leq \frac{\pi }{4} and  L_{2}  is the least value of the term independent of x when \frac{\pi }{16}\leq \theta \leq \frac{\pi }{8}, then the ratio L_{2}:L_{1}  is equal to : 
Option: 1 16:1
Option: 2 8:1
Option: 3 1:8
Option: 4 1:16
 

General Term of Binomial Expansion\left(T_{r+1}\right)^{\mathrm{th}} \text { term is called as general term in }(x+y)^{n}\;\text{and general term is given by}

\mathrm{T}_{\mathrm{r}+1}=^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{\;x}^{\mathrm{n}-\mathrm{r}} \cdot \mathrm{y}^{\mathrm{r}}

Term independent of x: It means term containing x0,

 

Now,

\\\mathrm{T}_{\mathrm{r}+1}=^{16} \mathrm{C}_{\mathrm{r}}\left(\frac{\mathrm{x}}{\cos \theta}\right)^{16-\mathrm{r}}\left(\frac{1}{\mathrm{x} \sin \theta}\right)^{\mathrm{r}}\\\text{for r = 8 term is free from 'x' }\\\begin{aligned} &\mathrm{T}_{9}=^{16} \mathrm{C}_{8} \frac{1}{\sin ^{8} \theta \cos ^{8} \theta}\\ &\mathrm{T}_{9}=^{16} \mathrm{C}_{8} \frac{2^{8}}{(\sin 2 \theta)^{8}}\\ &\text { in } \theta \in\left[\frac{\pi}{8}, \frac{\pi}{4}\right], L_{1}=^{16} \mathrm{C}_{8} 2^{8} \end{aligned}

\\\because \text{Min value of L}_1\;\text{at }\theta=\pi/4\\\text { in } \theta \in\left[\frac{\pi}{16}, \frac{\pi}{8}\right], L_{2}=16 \mathrm{C}_{8} \frac{2^{8}}{\left(\frac{1}{\sqrt{2}}\right)^{8}}=^{16} \mathrm{C}_{8} \cdot 2^{8} \cdot 2^{4}\\\\\because \text{Min value of L}_2\;\text{at }\theta=\pi/8\\\frac{L_{2}}{L_{1}}=\frac{16 \mathrm{C}_{8} \cdot 2^{8} 2^{4}}{^{16} \mathrm{C}_{8} \cdot 2^{8}}=16

Correct option 1

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Posted by

avinash.dongre

The value of  \cos ^{3}\left ( \frac{\pi }{8} \right )\cdot \cos \left ( \frac{3\pi }{8} \right )+\sin ^{3}\left ( \frac{\pi }{8} \right )\cdot \sin \left ( \frac{3\pi }{8} \right )  is :
Option: 1 \frac{1}{4}
 
Option: 2 \frac{1}{2\sqrt{2}}
 
Option: 3 \frac{1}{2}
 
Option: 4 \frac{1}{\sqrt{2}}
 
 

 

 

Trigonometric Identities -

Trigonometric Identities-

These identities are the equations that hold true regardless of the angle being chosen.

 

\\\mathrm{\sin^2\mathit{t}+\cos^2\mathit{t}=1}\\\mathrm{1+\tan^2\mathit{t}=\sec^2\mathit{t}}\\\mathrm1+{\cot^2\mathit{t}=\csc^2\mathit{t}}\\\mathrm{\tan \mathit{t}=\frac{\sin \mathit{t}}{\cos \mathit{t}},\;\;\cot \mathit{t}=\frac{\cos\mathit{t}}{\sin\mathit{t}}}

-

 

 

 

Allied Angles (Part 1) -

Allied Angles (Part 1)

Two angles or numbers are called allied iff their sum or difference is a multiple of π/2   

  • sin (900 - θ) = cos (θ)

  • cos (900 - θ) = sin (θ)

  • tan (900 - θ) = cot (θ)

  • csc (900 - θ) = sec (θ)          

  • sec (900 - θ) = csc (θ)

  • cot (900 - θ) = tan (θ)

 

  • sin (900 + θ) = cos (θ)

  • cos (900 + θ) = - sin (θ)

  • tan (900 + θ) = - cot (θ)

  • csc (900 + θ) = sec (θ)          

  • sec (900 + θ) = - csc (θ)

  • cot (900 + θ) = - tan (θ)

-

 

 

 

\\\cos ^3\:\frac{\pi }{8}\cos \frac{3\:\pi }{8}+\sin ^3\frac{\pi }{8}\:\sin \frac{3\:\pi }{8}\\\sin\left ( \frac{\pi}{2}-\frac{3\pi}{8} \right )=\cos\left ( \frac{3\pi}{8} \right )=\sin\left ( \frac{\pi}{8} \right )\\\\\\\cos ^3\:\frac{\pi }{8}\sin \frac{\:\pi }{8}+\sin ^3\frac{\pi }{8}\:\cos \frac{\:\pi }{8}\\\sin\frac{\pi}{8}\cos\frac{\pi}{8}\left ( \cos ^2\:\frac{\pi }{8}+\sin ^2\:\frac{\pi }{8} \right )\\\frac{1}{2}\left ( 2\:\sin \frac{\pi }{8}\cos \:\frac{\pi \:}{8} \right )=\frac{1}{2}\:\sin \frac{2\pi }{8}=\frac{1}{2\sqrt2}

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Posted by

avinash.dongre

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At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion.  After combustion, the gases occupy 330 mL.  Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is :
Option: 1  C4H8  
Option: 2  C4H10
Option: 3  C3H6
Option: 4  C3H8
 

Volume of N2  in air = 375 × 0.8 = 300 ml

Volume of O2 in air = 375 × 0.2 = 75 ml
 

C_{x}H_{y} +\left ( x +\frac{y}{4} \right )O_{2} \; \rightarrow \; xCO_{2}(g) + \frac{y}{2} H_{2}O(l)

15ml                15\left ( x +\frac{y}{4} \right )

  0                         0                            15x                 -

 

After combustion total volume

330 =V_{N_{2}} + V_{CO_{2}}

330 = 300 + 15x 

x = 2 

Volume of O2 used

15\left ( x +\frac{y}{4} \right ) = 75

\left ( x +\frac{y}{4} \right ) = 5

y = 12 

So hydrocarbon is = C2H12

None of the options matches it therefore it is a BONUS.

----------------------------------------------------------------------

Alternatively  Solution


 C_{x}H_{y} +\left ( x +\frac{y}{4} \right )O_{2} \; \rightarrow \; xCO_{2}(g) + \frac{y}{2} H_{2}O(l)

15ml              15\left ( x +\frac{y}{4} \right )

  0                         0                            15x                 -

Volume of O2 used

15\left ( x +\frac{y}{4} \right ) = 75

\left ( x +\frac{y}{4} \right ) = 5

If further information (i.e., 330 ml) is neglected, option (C3H8 ) only satisfy the above equation.

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Posted by

Ritika Jonwal

 An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant.  If during  this process the relation of pressure P and volume V is given by PVn=constant,  then n is given by (Here CP and CV are molar specific heat at constant pressure and constant volume, respectively)
Option: 1  n=\frac{C_{p}}{C_{v}}


Option: 2  n=\frac{C-C_{p}}{C-C_{v}}


Option: 3 n=\frac{C_{p}-C}{C-C_{v}}

Option: 4  n=\frac{C-C_{v}}{C-C_{p}}
 

For a polytropic preocess

c=c_{v}+\frac{R}{1-n} \: or \: \frac{R}{1-n} = c-c_{v}

\Rightarrow 1-n=\frac{R}{c-c_{v}} \: or\: n=1-\frac{R}{c-c_{v}}

\Rightarrow n=\frac{c-\left ( c_{v}+R \right )}{c-c_{v}} = \frac{c-c_{p}}{c-c_{v}}

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Posted by

Ritika Jonwal

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