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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

 

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Posted by

DHFM DVP

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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}}=0.29where h=2(given)

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Posted by

Ritika Jonwal

 A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed \nu in the x-y plane as shown in the figure : Which of the following statements is false for the angular momentum  \vec{L} about the
origin?
Option: 1 \vec{L}= -\frac{mv}{\sqrt{2}}R\hat{k}when the particle is moving from A to B.

Option: 2 \vec{L}= mv\left [ \frac{R}{\sqrt{2}}-a \right ]\hat{k}when the particle is moving from C to D.

Option: 3 \vec{L}= mv\left [ \frac{R}{\sqrt{2}}+a \right ]\hat{k}when the particle is moving from B to C.

Option: 4 L=\frac{R}{\sqrt{2}} m v(-k)when the particle is moving from D to A.
 

\\ In\ option\ (a), co-ordinates \ of\ A are \left(\frac{R}{\sqrt{2}}, \frac{R}{\sqrt{2}}\right) \\ \therefore \vec{r}=\left(\frac{R}{\sqrt{2}} \hat{i}+\frac{R}{\sqrt{2}} \hat{j}\right) and \vec{v}=v \hat{i}\\ \vec{L} m(\vec{r} \times \vec{v})=m\left(\frac{R}{\sqrt{2}} \hat{i}+\frac{R}{\sqrt{2}} \hat{j}\right) \times v \hat{i}\\ \vec{L}=-\frac{m R}{\sqrt{2}} v \hat{k}

\\ in\ option \ (b)\ it \ moves\ from \ C \ to\ D\\ L=\left(\frac{R}{\sqrt{2}}+a\right) m v(\hat{k})

so option b is correct option

\\ in \ option\ (c), \ For\ B$ \ to\ $C,$ \\ we have $L=\left(\frac{R}{\sqrt{2}}+a\right) m v(\hat{k})

\\ in \ option \ (d), \ When \ a \ particle\ is \ moving\ from \ D \ to\ A\\ L=\frac{R}{\sqrt{2}} m v(-k)

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Posted by

Ritika Jonwal

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The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight (in kg) which a 75 kg person would gain if all ^{1}H atoms are replaced by ^{2}H atoms is :  
Option: 1 7.5
Option: 2 10
Option: 3 15
Option: 4 37.5
 

Given that

Mass of the person = 75 kg

Mass of 1H1 present in person = 10% of 75 kg = 7.5 kg

Since Mass of 1H2 is double the Mass of 1H1

So, Mass of 1H2 will be in person = 2 X 7.5 kg =15 kg

Thus, increase in weight = 15 - 7.5 = 7.5 kg

Therefore, Option (1) is correct

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Posted by

vishal kumar

A body of mass m=10−2 kg is moving in a medium and experiences a frictional forceF= -kv^{2}  Its initial speed is v_{0}= 10ms^{-1}.  If, after 10 s, its energy is \frac{1}{8}  mv{_{0}}^{2}  the value of k will be :  
Option: 1 10−3 kg m−1  
Option: 2  10−3 kg s−1  
Option: 3  10−4 kg m−1  
Option: 4 10−1 kg m−1 s−1  
 

As we learnt in 

\frac{1}{2}mv{_{f}}^{2}=\frac{1}{8}mv{_{o}}^{2}

v_{f}=\frac{Vo}{2}=5 m \slash s

\left ( 10^{-2} \right )\frac{dv}{dt}=-KV^{2}

\int_{10}^{5}=-100k\int_{0}^{10}dt\Rightarrow \frac{1}{5}-\frac{1}{10}=100 k (1^{\circ})

K=10^{-4}

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Posted by

vishal kumar

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The correct order of the atomic radii of C, Cs, Al and S is :
Option: 1 S<C<Cs<Al
Option: 2 C<S<Al<Cs
Option: 3 C<S<Cs<Al
Option: 4 S<C<Al<Cs
 

 

Periodicity of atomic radius and ionic radius in period -

In a period from left to right the effective nuclear charge increases because the next electron fills in the same shell. So the atomic size decrease.

- wherein

Li>Be>B>C>N>O>F

 

 

 

Electronegativity and atomic radius -

The attraction between the outer electrons and the nucleus increases as the atomic radius decreases in a period.

- wherein

Electronegativity\propto\frac{1}{atomic\:radius}

 

 

 

Size of atom and ion in a group -

In a group moving from top to the bottom the number of shell increases.So the atomic size increases.

- wherein

Li<Na<K<Rb<Cs

As we know that

From Left to right in a period size decreases and when going down the group size increases

C< S< Al< Cs

Therefore, Option(2) is correct

  

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Posted by

Ritika Jonwal

The IUPAC symbol for the element with atomic number 119 would be:
Option: 1 uue
Option: 2une
Option: 3 unh
Option: 4 uun

 

Nomenclature of elements with atomic number >100 -

The name is derived directly from the atomic number of the element using the following numerical roots:

0 = nil

1 = un

2 = bi

3 = tri

4 = quad

5 = pent

6 = hex

7 = sept

8 = oct

9 = enn

Eg:

 

Atomic number

Name

Symbol

101

Mendelevium (Unnilunium)

Md (Unu)

102

Nobelium (Unnilbium)

No (Unb)



 

-

uue

1  1  9

Un Un ennium

Therefore, Option(1) is correct.

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Posted by

Ritika Jonwal

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Consider a uniform rod of massM=4m and length l pivoted about its centre. A mass m moving with velocity v making angle \theta =\frac{\pi }{4} to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod - mass system just after the collision is :   
Option: 1 \frac{4}{7}\frac{v}{l}

Option: 2 \frac{3\sqrt{2}}{7}\frac{v}{l}

Option: 3 \frac{3}{7\sqrt{2}}\frac{v}{l}  

Option: 4 \frac{3}{7}\frac{v}{l}

 

Let us conserve angular momentum about O:-

           So, L_i=\left ( \frac{mv}{\sqrt2} \right )\times \frac{l}{2}, where \left ( \frac{mv}{\sqrt2} \right ) is linear momentum and \left ( \frac{l}{2} \right ) is the distance from centre O.

Now, L_f=I\omega

Here, I=\frac{4ml^2}{12}+{m\left (\frac{l}{2} \right )^2}=\frac{7ml^2}{12}

So, L_i=L_f\Rightarrow \omega=\frac{6v}{7\sqrt2 l}= \frac{3\sqrt2v}{7 l}

So the correct graph is given in option 2.

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Posted by

vishal kumar

A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length l. The other end is fixed. The system is given an angular speed \omega about the fixed end of the spring such that it rotates in a circle in gravity-free space. Then the stretch in the spring is :     
Option: 1 \frac{ml\omega ^2}{k-\omega m}
 
Option: 2 \frac{ml\omega ^2}{k-m\omega ^2}

Option: 3 \frac{ml\omega ^2}{k+m\omega ^2}  

Option: 4 \frac{ml\omega ^2}{k+m\omega }
 

 

 

 

 

As natural lentgh=l

Let elongation=x

Mass m is moving with angular velocity \omega in a radius r

where r=l+x

Due to elongation x spring force is given by F_s=Kx

And F_C=m\omega ^2r=m\omega ^2(l+x)

as F_C=F_s

So

 Kx =m\omega ^2(l+x)\\ \Rightarrow x=\frac{m\omega ^2l}{K-m\omega ^2}

So the correct option is 2.

 

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Posted by

vishal kumar

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