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

Not understanding sir

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

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

#### The value of    is : Option: 1   Option: 2   Option: 3   Option: 4

Trigonometric Identities -

Trigonometric Identities-

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

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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)

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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 N 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.

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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.

#### 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

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#### A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure.  The coefficient of friction, between the particle and the rough track equals µ.  The particle is released, from rest, from the point P and it comes to rest at a point R.  The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction µ and the distance x(=QR), are, respectively close to : Option: 1  0.2 and 6.5 m   Option: 3  0.2 and 3.5 m   Option: 4   0.29 and 6.5 m

Work done by friction at QR = μmgx

In triangle, sin 30° = 1/2 = 2/PQ

PQ = 4 m

Work done by friction at PQ = μmg × cos 30° × 4 = μmg × √3/2 × 4 = 2√3μmg

Since work done by friction on parts PQ and QR are equal,

μmgx = 2√3μmg

x = 2√3 ≅ 3.5 m

Applying work energy theorem from P to R

decrease in P.E.=P.E.= loss of energy due to friction in PQPQ and QR

$\\ m g h=(\mu m g \cos \theta) P Q+\mu m g \times Q R\\ h=\mu \cos \theta \times P Q+\mu m g \times Q R\\ h=\mu \cos \theta \times P Q+\mu \times Q R =\mu \cos 30^{\circ} \times 4+\mu \times 2 \sqrt{3} =\mu\left(4 \times \frac{\sqrt{3}}{2}+2 \sqrt{3}\right)\\ h=\mu \times 4 \sqrt{3}\\ \mu=\frac{2}{4 \sqrt{3}}=\frac{1}{2 \sqrt{3}}=0.29$where h=2(given)

#### In the following structure, the double bonds are marked as I, II, III and IV Geometrical isomerism is not possible at site (s) : Option: 1  III Option: 2  I Option: 3  I and II Option: 4  III and IV

Geometrical isomerism is not possible at Site I as two identical methyl groups are attached to the same carbon bearing the double bond.

Hence, the answer is Option (2)