Filter By

## All Questions

#### 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

1.32

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

2

#### 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.

## JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

#### The following system of linear equations   has  Option: 1 infinitely many solutions,  satisfying  Option: 2 infinitely many solutions,  satisfying  Option: 3 no solution Option: 4 only the trivial solution.

System of Homogeneous linear equations -

If ? ≠ 0, then x= 0, y = 0, z = 0 is the only solution of the above system. This solution is also known as a trivial solution.

If ? = 0, at least one of x, y and z are non-zero. This solution is called a non-trivial solution.

Explanation: using equation (ii) and (iii), we have

This is the condition for a system have Non-trivial solution.

-

so infinite non-trivial solution exist

now equation (1) + 3 equation (3)

10x - 20z = 0

x = 2z

Correct Option 2

#### Let   If  then :    Option: 1 Option: 2 Option: 3 Option: 4

Elementary row operations -

Elementary row operations

Row transformation: Following three types of operation (Transformation) on the rows of a given matrix are known as elementary row operation (transformation).

i) Interchange of ith row with jth row, this operation is denoted by

ii) The multiplication of ith row by a constant k (k≠0) is denoted by

iii) The addition of ith row to the elements of jth row multiplied by constant k (k≠0) is denoted by

In the same way, three-column operations can also be defined too.

-

Correct Option (3)

## NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

#### If for some  in R, the intersection of the following three planes  is aline in , then  is equal to :  Option: 1 Option: 2 Option: 3  Option: 4

Cramer’s law -

Cramer’s law for the system of equations in two variables :

We can observe that first row in the numerator of x is of constants and 2nd row in the numerator is of constants, and the denominator is of the coefficient of variables.

We can follow this analogy for the system of equations of 3 variable where third in the numerator of the value of z will be of constant and denominator will be formed by the value of coefficients of the variables.

i) If ? ≠ 0, then the system of equations has a unique finite solution and so equations are consistent, and solutions are

ii) If ? = 0, and any of

Then the system of equations is inconsistent and hence no solution exists.

iii) If all  then

System of equations is consistent and dependent and it has an infinite number of solution.

-

#### 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.

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

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.

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

#### 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

#### 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)