Q&A - Ask Doubts and Get Answers

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

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

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

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

Guru G

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

Guru G

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

A long wire having cross-sectional area and Young's modulus is subjected to a load of , then the elongation in the wire will be: Option: 1 Option: 2 Option: 3 Option: 4

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

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 $3 \sqrt{2}: 2$

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 $2$