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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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Posted by

Raju vittal nandi

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

1.32

View Full Answer(4)

Posted by

balda gayathri

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 G₂M and M

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

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Posted by

Ajit Kumar Dubey

The following system of linear equations $7x+6y-2z=0$ $3x+4+2z=0$ $x-2y-6z=0,$ has
Option: 1 infinitely many solutions, $(x,y,z)$ satisfying $y=2z.$
Option: 2 infinitely many solutions, $(x,y,z)$ satisfying $x=2z.$
Option: 3 no solution
Option: 4 only the trivial solution.

System of Homogeneous linear equations -

$\\\mathrm{Let,} \\\mathrm{a_1x+b_1y +c_1z=0\;\;\; ...(i)} \\\mathrm{a_2x+b_2y +c_2z=0\;\;\; ...(ii)} \\\mathrm{a_3x+b_3y +c_3z=0\;\;\; ...(iii)} \\\mathrm{be \; three\; homogeneous\; equation} \\\mathrm{and \; let\; \Delta = \begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix}}$

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

$\\\mathrm{\frac{x}{b_2c_3-b_3c_2} = \frac{y}{c_2a_3-c_3a_2}=\frac{z}{a_2b_3-a_3b_2}} \\\\\mathrm{or \;\; \frac{x}{\begin{vmatrix} b_2 &c_2 \\ b_3 & c_3 \end{vmatrix}}=\frac{y}{\begin{vmatrix} c_2 &a_2 \\ c_3 & a_3 \end{vmatrix}} = \frac{z}{\begin{vmatrix} a_2 & b_2\\ a_3 & b_3 \end{vmatrix}} = k (say \neq 0)} \\\mathrm{\therefore x = k\begin{vmatrix} b_2 &c_2 \\ b_3 & c_3 \end{vmatrix}, y = k\begin{vmatrix} c_2& a_2\\ c_3 & a_3 \end{vmatrix} \; and \; z = k\begin{vmatrix} a_2 & b_2\\ a_3 & b_3 \end{vmatrix}} \\\mathrm{putting\; these\; value\; in \; equation \; (i), we\; have} \\\mathrm{a_1\left \{ k\begin{vmatrix} b_2 & c_2\\ b_3 & c_3 \end{vmatrix} \right \}+b_1\left \{ k\begin{vmatrix} c_2 & a_2\\ c_3 & a_3 \end{vmatrix} \right \}+c_1\left \{ \begin{vmatrix} a_2 & b_2\\ a_3 & b_3 \end{vmatrix} \right \}=0}$

$\\\mathrm{\Rightarrow a_1\begin{vmatrix} b_2 &c_2 \\ b_3 & c_2 \end{vmatrix}-b_1\begin{vmatrix} a_2 & c_2\\ a_3 &c_3 \end{vmatrix}+c_1\begin{vmatrix} a_2 &b_2 \\ a_3 &b_3 \end{vmatrix} = 0 \;\;\;[\because \; k \neq 0]} \\\mathrm{or \;\; \begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix} = 0 \; or \; \Delta = 0}$

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

-

$\begin{aligned} &(1)\;\;7 x+6 y-2 z=0\\ &(2)\;\;3 x+4 y+2 z=0\\ &(3)\;\;x-2 y-6 z=0 \end{aligned}$

$\left|\begin{array}{ccc}{7} & {6} & {-2} \\ {3} & {4} & {2} \\ {1} & {-2} & {-6}\end{array}\right| =7(-20)-6(-20)-2(-10)=-140+120+20=0$

so infinite non-trivial solution exist

now equation (1) + 3 equation (3)

10x - 20z = 0

x = 2z

Correct Option 2

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Posted by

avinash.dongre

Let $a-2b+c=1.$ If $f\left ( x \right )= \begin{vmatrix} x+a & x+2 &x+1 \\ x+b &x+3 &x+2 \\ x+c &x+4 & x+3 \end{vmatrix},$ then :
Option: 1 $f(-50)=501$
Option: 2 $f(-50)=-1$
Option: 3 $f(50)=1$
Option: 4 $f(50)=-501$

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 i^th row with j^th row, this operation is denoted by

$\\\mathrm{R_i \leftrightarrow R_j \;\;or\; R_{ij}}$

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

$\\\mathrm{R_i \leftrightarrow kR_i \;\; or \; k\cdot R_i}$

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

$\\\mathrm{R_i \leftrightarrow R_i + kR_j \;\; or \; k\cdot R_{ij}}$

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

-

$\\\text { Apply } \mathrm{R}_{1}=\mathrm{R}_{1}+\mathrm{R}_{3}-2 \mathrm{R}_{2}\\\Rightarrow f(x)=\left|\begin{array}{ccc}{1} & {0} & {0} \\ {x+b} & {x+3} & {x+2} \\ {x+c} & {x+4} & {x+3}\end{array}\right| \quad \Rightarrow f(x)=1 \\\quad \Rightarrow f(50)=1$

Correct Option (3)

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Posted by

avinash.dongre

If for some $\alpha \: and\: \beta$ in R, the intersection of the following three planes $x+4y-2z=1$ $x+7y-5z=\beta$ $x+5y+\alpha z=5$ is aline in $R^{3}$ , then $\alpha +\beta$ is equal to :
Option: 1 $0$
Option: 2 $10$
Option: 3 $-10$
Option: 4 $2$

Cramer’s law -

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

$\\\mathrm{Let \; a_1x +b_1y = c_1\; and \; a_2x + b_2y = c_2, where} \\\mathrm{\frac{a_1}{a_2}\neq\frac{b_1}{b_2}} \\\mathrm{on \; solving \; this \; equation \; by \; cross \; multiplication, we \; get} \\\mathrm{\frac{x}{b_1c_2-b_2c_1}=\frac{y}{a_1c_2-a_2c_1}=\frac{1}{a_1b_2-a_2b_1}} \\\mathrm{or \; \frac{x}{\begin{vmatrix} b_1 & c_1\\ b_2 & c_2 \end{vmatrix}}=\frac{y}{\begin{vmatrix} a_1 & c_1\\ a_2 & c_2 \end{vmatrix}}=\frac{1}{\begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}}} \\\mathrm{or \; x=\frac{\begin{vmatrix} b_1 & c_1\\ b_2 & c_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}}, y=\frac{\begin{vmatrix} a_1 & c_1\\ a_2 & c_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}}}$

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.

$\\\mathrm{so\; let \;us\; consider\; the \; system\; of \;equations} \\\mathrm{a_1x+b_1y +c_1z =d_1 \;\;...(i)} \\\mathrm{a_2x+b_2y +c_2z =d_2\;\;\;...(ii)} \\\mathrm{a_3x+b_3y +c_3z =d_3\;\;\;...(iii)} \\\mathrm{then \; \Delta,\; which\; will\; be \;determinant\; of\; coefficient\; of \;variables, will\; be } \\\mathrm{\Delta = \begin{vmatrix} a_1 & b_1 & c_1\\ a_2& b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix}, \Delta_1\; numerator\; of\; x \;is:} \\\mathrm{\Delta_1= \begin{vmatrix} d_1 & b_1 &c_1 \\ d_2 & b_2 & c_2\\ d_3 & b_3 & c_3 \end{vmatrix}, similarly\; \Delta_2 = \begin{vmatrix} a_1 & d_1 & c_1\\ a_2 & d_2 & c_2\\ a_3 & d_3 & c_3 \end{vmatrix}\; and \;} \\\mathrm{\Delta_3 = \begin{vmatrix} a_1 & b_1 & d_1\\ a_2 & b_2 & d_2\\ a_3 & b_3 & d_3 \end{vmatrix}, if \Delta \neq 0\; then\; \;x=\frac{\Delta_1}{\Delta}} \\\mathrm{y=\frac{\Delta_2}{\Delta}, z=\frac{\Delta_3}{\Delta}}$

i) If ? ≠ 0, then the system of equations has a unique finite solution and so equations are consistent, and solutions are $\\\mathrm{x=\frac{\Delta_1}{\Delta}, y=\frac{\Delta_2}{\Delta}, z=\frac{\Delta_3}{\Delta}}$

ii) If ? = 0, and any of $\Delta_1\neq 0 \; or \;\Delta_2\neq 0 \; or \;\Delta_3\neq 0$

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

iii) If all $\Delta =\Delta_1=\Delta_2=\Delta_3= 0$ then

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

-

$\begin{array}{l}{\Delta=0 \Rightarrow\left|\begin{array}{ccc}{1} & {4} & {-2} \\ {1} & {7} & {-5} \\ {1} & {5} & {\alpha}\end{array}\right|=0} \\ {(7 \alpha+25)-(4 \alpha+10)+(-20+14)=0} \\ {3 \alpha+9=0 \Rightarrow \alpha=-3}\end{array}$

$\begin{array}{l}{\text { Also }\quad D_{3}=0 \Rightarrow\left|\begin{array}{lll}{1} & {4} & {1} \\ {1} & {7} & {\beta} \\ {1} & {5} & {5}\end{array}\right|=0} \\ {1(35-5 \beta)-(15)+1(4 \beta-7)=0} \\ {\beta=13}\end{array}$

$\alpha+ \beta=10$

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Posted by

avinash.dongre

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 O₂used

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

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

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 C_P and C_V 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

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)

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Posted by

Raju vittal nandi

Calculate the acceleration of block of the following diagram. Assume all surfaces are frictionless . Here m1 = 100kg and m2 = 50kg Option: 1 0.33m/s2Option: 2 0.66m/s2Option: 3 1m/s2Option: 4 1.32m/s2

Posted by

balda gayathri

Crack CUET with india's "Best Teachers"

No. of transition state in given figure Option: 1 1Option: 2 2Option: 3 3Option: 4 4

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

Posted by

Ajit Kumar Dubey

JEE Main high-scoring chapters and topics

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.

Posted by

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

No. of transition state in given figure

Option: 1
1

Option: 2
2

Option: 3
3

Option: 4
4

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 G₂M and M

The following system of linear equations $7x+6y-2z=0$ $3x+4+2z=0$ $x-2y-6z=0,$ has
Option: 1 infinitely many solutions, $(x,y,z)$ satisfying $y=2z.$
Option: 2 infinitely many solutions, $(x,y,z)$ satisfying $x=2z.$
Option: 3 no solution
Option: 4 only the trivial solution.

Let $a-2b+c=1.$ If $f\left ( x \right )= \begin{vmatrix} x+a & x+2 &x+1 \\ x+b &x+3 &x+2 \\ x+c &x+4 & x+3 \end{vmatrix},$ then :
Option: 1 $f(-50)=501$
Option: 2 $f(-50)=-1$
Option: 3 $f(50)=1$
Option: 4 $f(50)=-501$

If for some $\alpha \: and\: \beta$ in R, the intersection of the following three planes $x+4y-2z=1$ $x+7y-5z=\beta$ $x+5y+\alpha z=5$ is aline in $R^{3}$ , then $\alpha +\beta$ is equal to :
Option: 1 $0$
Option: 2 $10$
Option: 3 $-10$
Option: 4 $2$