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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

The following is the biochemical pathway for purple pigment production in flowers of sweet pea:

Colorless precursor 1 —- Allele A—-> Colorless precursor 2 —--- Allele B—---> Purple pigment

Recessive mutation of either gene A or B leads to the formation of white flowers. A cross is made between two parents with the genotype: AaBb × aabb. Considering that the two genes are not linked, the phenotypes of the expected progenies are:

Option: 1
9 purple : 7 white

Option: 2
3 white : 1 purple

Option: 3
1 purple : 1 white

Option: 4
9 purple : 6 light purple : 1 white

3(white):1(purple)

In case of 'duplicate recessive epistasis homozygous recessive gene masks the other gene i.e., aa is epistatic to A or a || bb is epistatic to B or b

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Posted by

Riya

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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

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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

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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 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 random variable X has the following probability distribution: $X: \: \: 1\; \; \: \: 2\; \;\: \: \: \: 3\; \; \: \: \: \: \: 4\: \: \: \; \; \: \: 5$ $P(X): \;K^{2}\; \; 2K\; \; K\; \; 2K\; \; 5K^{2}$ Then $P(X>2)$ is equal to:
Option: 1 $\frac{7}{12}$
Option: 2 $\frac{23}{36}$
Option: 3 $\frac{1}{36}$
Option: 4 $\frac{1}{6}$

Option 2

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Posted by

kpvalli2725@gmail.com

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Let $a_{n}$ be the n^th term of a G.P. of positive terms. If $\sum_{n=1}^{100}a_{2n+1}=200\: \: and\: \: \sum_{n=1}^{100}a_{2n}=100,\: \: then\: \sum_{n=1}^{200}a_n$ is equal to :
Option: 1 $300$
Option: 2 $175$
Option: 3 $225$
Option: 4 $150$

150

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Posted by

Nalla mahalakshmi

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

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If $x=\sum_{n=0}^{\infty }(-1)^{n}\tan ^{2n}\theta \: \: and\: \: y=\sum_{n=0}^{\infty }\cos ^{2n}\theta ,$ for $0<\theta < \frac{\pi }{4},$ then
Option: 1 $y(1+x)=1$
Option: 2 $x(1-y)=1$
Option: 3 $y(1-x)=1$
Option: 4 $x(1+y)=1$

$x=\sum_{n=0}^{\infty}(-1)^{n} \tan ^{2 n} \theta=1-\tan^2\theta+\tan^4\theta..........$

$y=\sum_{n=0}^{\infty} \cos ^{2 n} \theta=1+\cos^2\theta+\cos^4\theta......$

Use $\text S_{\infty}=\frac{1}{1-r}$

${x=\frac{1}{1+\tan ^{2} \alpha}=\cos ^{2} \theta} \\ {y=\frac{1}{1-\cos ^{2} \theta}=\frac{1}{\sin ^{2} \theta}}$

$\Rightarrow (1-x)= \sin ^{2} \theta$

$\Rightarrow y(1-x)=1$

Correct Option (3)

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Posted by

avinash.dongre

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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