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Villis are present in small intestine not in stomach because stomach doesn't absorb nutrients.

Villi of the small intestine are small finger-like projections on the surface. One villus contains many cells and thereby increase the surface area. This is required because a lot of nutrients have to be absorbed in the small intestine. Lots of cells on a small area makes this process efficient as, stomach collects the food, adds gastric acid and enzymes for digestion of food. The stomach does not absorb any nutrients and therefore does not require a larger surface area.

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Posted by

Priyanka Kumari

For the reaction:

$X_{2}O_{4}\left ( l \right )\rightarrow 2XO_{2}(g)$

$\Delta U=2.1k \ cal, \Delta S=20 cal\ K^{-1}$ at 300K

Hence $\Delta G$ is :-

Option 1)
2.7k cal

Option 2)
-2.7 k cal

Option 3)
9.3k cal

Option 4)
-9.3k cal

Molar heat capacity for isobaric process C(p) -

$dH=n\: C_{P}\: dT$

- wherein

$dH=dE+d(PV)$

or

$dH=dE+n\, R\, dT$

$\Delta H= \Delta U + \Delta n_{g}RT$

Given, $\Delta U=2.1 Kcal\;\;\;\; \Delta n_{g}=2$

$R- 2 \times 10^{-3}Kcal;\;\;T=300 K$

$\therefore \Delta H= 2.1+2 \times 2 \times 10^{-3} \times 300=3.3 Kcal$

Also, $\Delta G= \Delta H-T \Delta S$

Given, $\Delta S=20 \times 10^{-3}Kcal\;K^{-1}$

On putting the value of $\Delta H$ and $\Delta S$ in the equation, we get -

$\Delta G = 3.3-300 \times 20 \times 10^{-3}= -2.7 Kcal$

Option 1)

2.7k cal

This option is incorrect.

Option 2)

-2.7 k cal

This option is correct.

Option 3)

9.3k cal

This option is incorrect.

Option 4)

-9.3k cal

This option is incorrect.

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Posted by

Plabita

A thermodynamic system undergoes cyclic process ABCDA as shown in fig. the work done by the system in the cycle is:

Option 1)
$P_{o}V_{o}$

Option 2)
$2P_{o}V_{o}$

Option 3)
$\frac{P_{o}V_{o}}{2}$

Option 4)
$Zero$

Using

Cyclic Process -

A cyclic process consist of a number of changes that return the system back to its initial state.

- wherein

in cyclic process $Delta U= 0$

Work done in cyclic process = Area of the cycle

In the upper half, cycle is anti-clockwise while in lower half it is clockwise. Hence, net work done is zero.

Option 1)

$P_{o}V_{o}$

This option is incorrect

Option 2)

$2P_{o}V_{o}$

This option is incorrect

Option 3)

$\frac{P_{o}V_{o}}{2}$

This option is incorrect

Option 4)

$Zero$

This option is correct

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Posted by

divya.saini

A monoatomic gas at a pressure P, having a volume V expands isothermally to a volume 2 V, and then adiabatically to a volume 16 V. the final pressure of the gas is:

$take\:\:\left( \gamma= \frac{5}{3} \right )$

Option 1)
64 P

Option 2)
32 P

Option 3)
$\frac{P}{64}$

Option 4)
16 P

As learnt in

Equation of state -

$dQ= 0$

$n, C_{V}, dT+PdV= 0$

- wherein

On solving

$gamma frac{dV}{V}+frac{dP}{P}= 0$

$Rightarrow PV^{gamma }= constant$

For Isothermal expansion,

$P_{i}= P, \ V_{i}= V,\ V_{1}= 2V\ \ \Rightarrow P_{1} =P_{i}\cdot \frac{V}{2V}= \frac{P}{2}$

For adiabatic expansion,

$P_{1}= \frac{P}{2},\ V_{1}=2V,\ V_{f}= 16V,\ P_{f} =?$

$PV^{\gamma }= \ constant$

$\Rightarrow P_{1}V_{1}^{\gamma } = P_{f}V_{f}^{\gamma }\ \ \Rightarrow P_{f}= P_{1}\cdot \left ( \frac{V_{1}}{V_{f}} \right )^{\gamma }$

$P_{f}= \frac{P}{2}\cdot \left ( \frac{2V}{16V} \right )^{\frac{5}{3}} = \frac{P}{2}\cdot \left ( \frac{1}{2} \right )^{5} = \frac{P}{64}$

Option 1)

64 P

This option is incorrect

Option 2)

32 P

This option is incorrect

Option 3)

$\frac{P}{64}$

This option is correct

Option 4)

16 P

This option is incorrect

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Posted by

prateek

A body of mass (4m) is lying in x-y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (m) move perpendicular to each other with equal speeds (V). The total Kinetic energy generated due to explosion is

Option 1)
$mV^{2}$

Option 2)
$\frac{3}{2} mV^{2}$

Option 3)
$2 mV^{2}$

Option 4)
$4 mV^{2}$

seems logical .we assume that the 3rd part ie mass 2m goes in (-x-y) plane though the problem does not specify.How to explain that mass 2m does not move in Z direction?

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Posted by

v muralidharan

Steam at 100^o C is passed into 20 g of water at 10^o C. When water acquires a temperature of 80^o C, the mass of water present will be:

[Take specific heat of water = 1 cal g^-1 ^oC^-1 and latent heat of steam = 540 cal g^-1]

Option 1)
24 g

Option 2)
31.5 g

Option 3)
42.5 g

Option 4)
22.5 g

As discussed in

Specific Heat -

Amount of heat(Q) required to raise the temperature of unit mass through 1^oC.

- wherein

$Units-Calories/gm imes ^{o}C$

Specific heat of water $S_{w} = 1 \:\:cal\:\: g^{-1} \:\: ^{\circ} C^{-1}$

Latent heat of steam $L_{s}= 540 \:\: cal \:\: g^{-1}$

$Q_{1} = mL_{s}+mS_{w} \Delta T_{w} = m\times 540+m\times 1\times \left | 100-80 \right |$

$= 540m+20m = 560 m$

Heat gained by 20g of water to change its temperature from $10^{\circ}C$ to $80^{\circ}C$

$Q_{2} = m_{w}S_{w}\Delta T_{w} = 20\times 1\times \left | 80-10 \right | = 1400$

According to principle of calorimetry $Q_{1} = Q_{2}$

$\therefore 5600w = 1400$ or $w = 2.5g$

total mass of water = $= (20+m)g = (20+2.5)g= 22.5 g$

Option 1)

24 g

This solution is incorrect

Option 2)

31.5 g

This solution is incorrect

Option 3)

42.5 g

This solution is incorrect

Option 4)

22.5 g

This solution is correct

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Posted by

Vakul

Copper of fixed volume 'V' is drawn into wire of length 'l'. When this wire is subjected to a constant force'F', the extension produced in the wire is $'\Delta l'$ . Which of the following graphs is a straight line?

Option 1)
$\Delta l \:versus\:\frac{1}{l}$

Option 2)
$\Delta l \:versus\:l^{2}$

Option 3)
$\Delta l \:versus\:\frac{1}{l^{2}}$

Option 4)
$\Delta l \:versus\:l$

As discussed in

Young Modulus -

Ratio of normal stress to longitudnal strain

it denoted by Y

$Y= frac{Normal : stress}{longitudnal: strain}$

- wherein

$Y=frac{F/A}{Delta l/L}$

F - applied force

A - Area

$Delta l$ - Change in lenght

l - original length

$v = Al$

$Y = \frac{F/A}{\Delta l/l} = \frac{Fl}{A.\Delta L} \Rightarrow \Delta l = \frac{Fl}{YA}$

$\Delta l = \frac{Fl^{2}}{YV} \Rightarrow \Delta l \propto l^{2}$

Option 1)

$\Delta l \:versus\:\frac{1}{l}$

This solution is incorrect

Option 2)

$\Delta l \:versus\:l^{2}$

This solution is correct

Option 3)

$\Delta l \:versus\:\frac{1}{l^{2}}$

This solution is incorrect

Option 4)

$\Delta l \:versus\:l$

This solution is incorrect

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Posted by

Vakul

A certain number of spherical drops of a liquid of radius 'r' coalesce to form a single drop of radius 'R' and volume 'V'. If 'T' is the surface tension of the liquid, then:

Option 1)
$energy =4VT \left( \frac{1}{r}- \frac{1}{R} \right )is\:\:released$

Option 2)

$energy =3VT \left( \frac{1}{r}+ \frac{1}{R} \right )is\:\:absorbed$

Option 3)
$energy =3VT \left( \frac{1}{r}- \frac{1}{R} \right )is\:\:released$

Option 4)
Energy is neither released nor absorbed.

As discussed

Surface Energy -

It is defined as the amount of work done in increasing the area of the liquid against surface tension.

-

As surface area decreases so energy is released energy released = $4 \pi R^{2}T \left [ n^{\frac{1}{3}}-1 \right ]$

$R= n^{\frac{1}{3}}r$

$= 4 \pi R^{2}T\left [ \frac{1}{r} - \frac{1}{R}\right ] = 3 v \left [ \frac{1}{r}-\frac{1}{R} \right ]$

Option 1)

$energy =4VT \left( \frac{1}{r}- \frac{1}{R} \right )is\:\:released$

This solution is incorrect

Option 2)

$energy =3VT \left( \frac{1}{r}+ \frac{1}{R} \right )is\:\:absorbed$

This solution is incorrect

Option 3)

$energy =3VT \left( \frac{1}{r}- \frac{1}{R} \right )is\:\:released$

This solution is correct

Option 4)

Energy is neither released nor absorbed.

This solution is incorrect

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Posted by

perimeter

If the focal length of objective lens is increased then magnefying power of:

Option 1)
microscope will increase but that of telescope decrease.

Option 2)
microscope and telescope both will increase.

Option 3)
microscope and telescope both will decrease.

Option 4)
microscope will decrease but that of telescope increase.

Compound Microscope -

$m= -frac{v_{o}}{u_{o}}cdot frac{D}{u_{e}}$

- wherein

$v_{o}, and , u_{o}$ is distance from objective.

$u_{e}$ Distance from eyepiece.

Maximum magnification $m= -frac{v_{o}}{u_{0}}left ( 1+frac{D}{f_{e}} ight )$

Astronomical Telescope -

$m= frac{-f_{o}}{f_{e}}left ( 1+frac{f_{e}}{D} ight )$

- wherein

$f_{o}$ = focal length of objective

$f_{e}=$ focal length of eyepiece

For microscope magnifying power $\propto \frac{1}{fo}$

For telescope magnifying power $\propto fo$

$\therefore$ magnifying power of microscope decreases and that of telescope increases

Option 1)

microscope will increase but that of telescope decrease.

This option is incorrect

Option 2)

microscope and telescope both will increase.

This option is incorrect

Option 3)

microscope and telescope both will decrease.

This option is incorrect

Option 4)

microscope will decrease but that of telescope increase.

This option is correct

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Posted by

subam

In the Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is K, ( $\lambda$ being the wave length of light used). The intensity at a point where the path difference is $\frac{\lambda }{4}$ , will be:

Option 1)
$K$

Option 2)
$\frac{K}{4}$

Option 3)
$\frac{K}{2}$

Option 4)
$Zero$

Resultant Intensity of two wave -

$I= I_{1}+I_{2}+2sqrt{I_{1}I_{2}}cos heta$

- wherein

$I_{1}=$ Intencity of wave 1

$I_{2}=$ Intencity of wave 2

$heta =$ Phase difference

at a point where path difference is $\lambda=>phase difference =2\pi$

$\therefore I_{R}=K-(\sqrt{I_{1}}+\sqrt{I_{2}})^{2}$

$\therefore I_{R}=K-(\sqrt{I_{1}}+\sqrt{I_{2}})^{2}$ $=>K=4 I-----------------(1)$

$(I_{1} = I_{2}=I)$

When path a difference is $\frac{\lambda}{4}$

$=>$ phase difference = $\frac{\pi}{2}$

$I_{R}=I_{1} + I_{2} = 2I=\frac{k}{2}$

Option 1)

$K$

This option is incorrect

Option 2)

$\frac{K}{4}$

This option is incorrect

Option 3)

$\frac{K}{2}$

This option is correct

Option 4)

$Zero$

This option is incorrect

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hoping for quick and best answer

Posted by

Priyanka Kumari

For the reaction: at 300K Hence is :- Option 1) 2.7k cal Option 2) -2.7 k cal Option 3) 9.3k cal Option 4) -9.3k cal

Posted by

Plabita

Crack CUET with india's "Best Teachers"

A thermodynamic system undergoes cyclic process ABCDA as shown in fig. the work done by the system in the cycle is: Option 1) Option 2) Option 3) Option 4)

Posted by

divya.saini

A monoatomic gas at a pressure P, having a volume V expands isothermally to a volume 2 V, and then adiabatically to a volume 16 V. the final pressure of the gas is: Option 1) 64 P Option 2) 32 P Option 3) Option 4) 16 P

Posted by

prateek

JEE Main high-scoring chapters and topics

A body of mass (4m) is lying in x-y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (m) move perpendicular to each other with equal speeds (V). The total Kinetic energy generated due to explosion is Option 1) Option 2) Option 3) Option 4)

For the reaction:

$X_{2}O_{4}\left ( l \right )\rightarrow 2XO_{2}(g)$

$\Delta U=2.1k \ cal, \Delta S=20 cal\ K^{-1}$ at 300K

Hence $\Delta G$ is :-

Option 1)
2.7k cal

Option 2)
-2.7 k cal

Option 3)
9.3k cal

Option 4)
-9.3k cal

A thermodynamic system undergoes cyclic process ABCDA as shown in fig. the work done by the system in the cycle is:

Option 1)
$P_{o}V_{o}$

Option 2)
$2P_{o}V_{o}$

Option 3)
$\frac{P_{o}V_{o}}{2}$

Option 4)
$Zero$

A monoatomic gas at a pressure P, having a volume V expands isothermally to a volume 2 V, and then adiabatically to a volume 16 V. the final pressure of the gas is:

$take\:\:\left( \gamma= \frac{5}{3} \right )$

Option 1)
64 P

Option 2)
32 P

Option 3)
$\frac{P}{64}$

Option 4)
16 P

Steam at 100^o C is passed into 20 g of water at 10^o C. When water acquires a temperature of 80^o C, the mass of water present will be:

[Take specific heat of water = 1 cal g^-1 ^oC^-1 and latent heat of steam = 540 cal g^-1]

Option 1)
24 g

Option 2)
31.5 g

Option 3)
42.5 g

Option 4)
22.5 g