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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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If \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{xy}{x^{2}+y^{2}};y(1)=1; then a value of x satisfying y(x)=e is :   
Option: 1 \sqrt{3}\: e
 
Option: 2 \frac{1}{2}\sqrt{3}\: e
 
Option: 3 \sqrt{2}\: e
 
Option: 4 \frac{e}{\sqrt{2}}
 
 

√3e

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Posted by

Nalla mahalakshmi

If z be a complex number satisfying \left | Re\left ( z \right ) \right |+\left | Im(z) \right |=4, then \left | z \right | cannot be : 
Option: 1 \sqrt{7}
 
Option: 2 \sqrt{\frac{17}{2}}
 
Option: 3 \sqrt{10}
 
Option: 4 \sqrt{8}
 
 

Option d

 

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Posted by

Shravani.D.K

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Let a,b\epsilon \textbf{R},a\neq 0 be such that the equation, ax^{2}-2bx+5=0 has a repeated root \alpha, which is also a root of the equation, x^{2}-2bx-10=0. If \beta is the other root of this equation, then \alpha ^{2}+\beta ^{2} is equal to:
Option: 1 24
Option: 2 25
Option: 3 26
Option: 4 28
 

 

 

Nature of Roots -

Let the quadratic equation is ax2 + bx + c = 0

D is the discriminant of the equation.

iii) if roots D = 0, then roots will be real and equal, then


\\\mathrm{x_1=x_2 = \frac{-b}{2a} } \\\mathrm{Then, \;\; ax^2+bx +c =a(x-x_1)^2 }

-

ax2 – 2bx + 5 = 0 having equal roots or D=0 and \alpha=\frac{b}{a}

(2b)^2=4\times5\times a\;\;\Rightarrow \;\;b^2=5a

Put \alpha=\frac{b}{a} in the second equation

{x^{2}-2 b x-10=0} \\ {\Rightarrow b^{2}-2 a b^{2}-10 a^{2}=0}

\\\Rightarrow 5 a-10 a^{2}-10 a^{2}=0 \\ \Rightarrow 20 a^{2}=5 a \\ \Rightarrow a=\frac{1}{4} \text { and } \mathrm{b}^{2}=\frac{5}{4} \\ \alpha^{2}= 20 \text { and } \beta^{2}=5 \\ \alpha^{2}+\beta^{2} \\ =5+20 \\ =25

Correct Option 2

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Posted by

avinash.dongre

The  number of real roots of the equation,  e^{4x}+e^{3x}-4e^{2x}+e^{x}+1=0  is :   
Option: 1 3
Option: 2 4
Option: 3 1
Option: 4 2
 

 

 

Transcendental function -

Transcendental functions:  the functions which are not algebraic are called transcendental functions. Exponential, logarithmic, trigonometric and inverse trigonometric functions are transcendental functions.

Exponential Function: function f(x) such that \mathrm{f(x)=a^x} is known as an exponential function.

\\\mathrm{base:\;\;a>0,a\neq1}\\\mathrm{domain:x\in \mathbb{R}}\\\mathrm{range:f(x)>0}

 

 

Logarithmic function:  function f(x) such that f\left ( x \right )= \log\: _{a}x is called logarithmic function 

\\\mathrm{base:\;\;a>0,a\neq1}\\\mathrm{domain:x>0}\\\mathrm{range:f(x)\in\mathbb{R}}
         

                    If a > 1                                                                               If a < 1

Properties of Logarithmic Function

\\\mathrm1.\;{\log_e(ab)=\log_ea+\log_eb}\\\mathrm{2.\;\log_e\left ( \frac{a}{b} \right )=\log_ea-\log_e b}\\\mathrm{3.\;\log_ea^m=m\log_ea}\\\mathrm{4.\;\log_aa=1}\\\mathrm{5.\;\log_{b^m}a=\frac{1}{m}\log_ba}\\\mathrm{6.\;\log_ba=\frac{1}{\log_ab}}\\\mathrm{7.\;\log_ba=\frac{\log_ma}{\log_mb}}\\\mathrm{8.\;a^{\log_am}=m}\\\mathrm{9.\;a^{\log_cb}=b^{\log_ca}}\\\mathrm{10.\;\log_ma=b\Rightarrow a=m^b}

-

 

 

 

Quadratic Equation -

The root of the quadratic equation is given by the formula:

 

\\\mathrm{x = \frac{-b \pm \sqrt{D}}{2a}}\\\\\mathrm{or} \\\mathrm{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}

 Where D is called the discriminant of the quadratic equation, given by D = b^2 - 4ac ,

 

-

 

Let e^{x}=t \in(0, \infty)

Now the equation 

\begin{array}{l}{t^{4}+t^{3}-4 t^{2}+t+1=0} \\ {t^{2}+t-4+\frac{1}{t}+\frac{1}{t^{2}}=0} \\ {\left(t^{2}+\frac{1}{t^{2}}\right)+\left(t+\frac{1}{t}\right)-4=0}\end{array}

Let \mathrm{t}+\frac{1}{\mathrm{t}}=\alpha

\begin{array}{l}{\left(\alpha^{2}-2\right)+\alpha-4=0} \\ {\alpha^{2}+\alpha-6=0} \\ {\alpha^{2}+\alpha-6=0}\end{array}

\alpha=-3,2

Only positive value possible so \alpha=2 \Rightarrow \quad \mathrm{e}^{x}+\mathrm{e}^{-\mathrm{x}}=2

x=0 is the only solution.

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Posted by

avinash.dongre

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