datavorous/entrance-exam-dataset huggingface dataset:
A wire of length 20 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is

Options:
A 102+33
B 53+3
C 103+23
D 52+3

The correct answer is:
103+23

Solution:
Let the wire is cut into 2 pieces of length y & 20-y

Therefore side of square will be =y4 and hexagon will be =20-y6

Area of square=y42

Area of hexagon=6×3420-y62

Total area=y216+320-y224

On differentiating,

A'y=2y16+2320-y-124

A'y=0 at y=4033+23

Length of side of regular hexagon=1620-y=1620-4033+23=103+23

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (27 Aug Shift 1)
A wire of length 28 cm is bent to form a circular sector, then the radius (in cm) of the circular sector such that the area of the circular sector is maximum is equal to

Options:
A 5
B 6
C 7
D 8

The correct answer is:
7

Solution:
2r+rθ=28
θ=28-2rr … 1
Area =A=12r2θ
A=12r228-2rr
A=1228r-2r2
dAdr=28-4r2=0⇒r=7cm
Also, d2Adr2r=7<0
So, for the area to be maximum, r=7cm

Topic: Mathematics, Application of Derivatives, JEE Main
A wire of length 30 cm, stretched between rigid supports, has it's nth and n+1th harmonics at 400 Hz and 450 Hz, respectively. If tension in the string is 2700 N, it's linear mass density is _____ kg m-1.

Options:

The correct answer is:
3

Solution:
Frequency of nth harmonic is nv2l=400⇒nv0.6=400 ...1

And frequency of n+1th harmonic is n+1v2l=450⇒n+1v0.6=450 ...2, here, v is speed of wave on string.

From equation 1, we have

n=400×0.6v

Putting in equation 2,

⇒0.6×400v+1v0.6=450

⇒400+v0.6=450

⇒v=30 m s-1

Speed of wave on string is stated as v=Tμ=30, here, μ is linear mass density.

So, μ=Tv2=2700900⇒μ=3 kg m-1

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2022 (27 Jul Shift 2)
A wire of length 36 m is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is k (meter), then 4π+1k is equal to

Options:

The correct answer is:
36

Solution:
Let the side of square be x and radius of circle be y.

4x+2πy=36 (given)

Area, A=x2+πy2

⇒A=x2+π36-4x2π2

⇒⇒A=x2+1π18-2x2

⇒dAdx=2x+2π18-2x-2

⇒dAdx=2πx+(36-4x)(-2)π

For critical points

dAdx=0

⇒2πx-72+8x=0

⇒x=36π+4

Now,

d2Adx2=2+8π

⇒d2Adx2>0

Hence, area is minimized.

y=18π+4
K [circumference of circle)=2πy⇒k=36π(π+4)
∴4+ππ×k=36

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (26 Aug Shift 1)
A wire of length 100 m is tightly wounded on a hollow tube of radius 5mm and length 1 m. A current of 1 A is flowing in the wire. Then magnetic field strength inside the tube will be :-

Options:
A 4 T

B 4 mT

C 40 mT
D 40 T

The correct answer is:
4 mT

Solution:
Field due to a solenoid = μ 0 ni
Given i = 1 A
n = 1 0 0 2 π × 5 × 1 0 - 3
∴ B=μ0ni=4π×10-7×100×12π×5×10-3=4 mT

Topic: Physics, Magnetic Effects of Current, JEE Main
A wire of length 314 cm carrying current of 14 A is bent to form a circle. The magnetic moment of the coil is A-m2. [Given π=3.14]

Options:

The correct answer is:
11

Solution:
Length of the wire l when bent becomes circumference of the loop, whose radius is given by R=l2π=3141002×3.14=12 m

Magnetic moment of the coil is M=IA

=14×πR2

=14×3.14×14

=10.99≈11.00 A-m2

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (29 Jul Shift 1)
A wire of length L=20 cm is bent into a semi-circular arc and the two equal halves of the arc are uniformly charged with charges +Q and -Q as shown in the figure. The magnitude of the charge on each half is Q=103ε0, where ε0 is the permittivity of free the space. The net electric field at the centre O is

Options:
A 25×103i^ N C-1
B 50×103i^ N C-1
C 25×103j^ N C-1
D 50×103j^ N C-1

The correct answer is:
25×103i^ N C-1

Solution:
L=πR
⇒R= L π = 20 100π   m= 1 5π  m
⇒ due to a charge arc, electric field at centre is given by
E= 2Kλ R sin θ 2
E1=E2=2k⋋Rsin⁡902 {⋋=QπR/2}
undefined
Component along j^ gets cancelled and
E net = 2 E 1
= 4KQ π R 2
=4×1103ϵ04πϵ0 πR2=4×103ϵ04π2ϵ0R2=102R2=10015π2=25×103
E→net=25×103 N C-1 i^

Topic: Physics, Electrostatics, JEE Main, JEE Main 2015 (11 Apr Online)
A wire of length L=20 cm is bent into a semi-circular arc and the two equal halves of the arc are uniformly charged with charges +Q and -Q as shown in the figure. The magnitude of the charge on each half is Q=103ε0, where ε0 is the permittivity of free the space. The net electric field at the centre O is

Options:
A 25×103i^ NC-1
B 50×103i^ NC-1
C 25×103j^ NC-1
D 50×103j^ NC-1

The correct answer is:
25×103i^ NC-1

Solution:
L=ÏR
âR= L Ï = 20 100Ï ââm= 1 5Ï âm
â due to a charge arc, electric field at centre is given by
E= 2KÎ» R sin Î¸ 2
E1=E2=2kâRsinâ¡902 {â=QÏR/2}
E 1 =E 2 = 2 2 âKQ Ï R 2
Component along j^ gets cancelled and
E net = 2 E 1
= 4KQ Ï R 2
=4Ã1103Ïµ04ÏÏµ0 ÏR2=4Ã103Ïµ04Ï2Ïµ0R2=102R2=10015Ï2=25Ã103
Eânet=25Ã103NC i^

Topic: Physics, Magnetic Properties of Matter, JEE Main
A wire of length L and 3 identical cells, of negligible internal resistance, are connected in series. Due to the current, the temperature of the wire is raised by ÎT in a time t. A number of N similar cells are now connected in series with a wire of the same material and area of cross-section but of length 2L. The temperature of the wire is raised by the same amount ÎT in the same time. Find the value of N.

Options:

The correct answer is:
6

Solution:
In the first case
3E2Rt=msΔT ... (ii)
When the length of the wire is doubled, resistance and mass both are doubled. Therefore, in the second case,
NE22Rt=2msΔT ... (ii)
Dividing Eq.(ii) by (i) we get
N218=2
or N2=36
or N=6

Topic: Physics, Current Electricity, JEE Main
A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by Î T in a time t. N number of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. The temperature of the wire is raised by the same amount Î T in the same time. The value of N is

Options:

The correct answer is:
6

Solution:
In the first case, 3 E 2 R t = m s Î T ...(i)
H = V 2 R t
When length of the wire is doubled, resistance and mass both are doubled.
Therefore, in the second case,
N E 2 2 R Â· t = 2 m s Î T ...(ii)
Dividing equation (ii) by (i), we get
N218=2 or N2=36 or N=6.

Topic: Physics, Current Electricity, JEE Main
A wire of length L and mass per unit length 6.0×10-3 kg m-1 is put under tension of 540 N. Two consecutive frequencies that it resonates at are: 420 Hz and 490 Hz. Then L in meters is :

Options:
A 2.1m
B 1.1m
C 8.1m
D 5.1m

The correct answer is:
2.1m

Solution:
Fundamental frequency =490-420=70 Hz
70=12LTμ
⇒70=12L5406×10-3
⇒L=12×7090×103=300140
⇒L≈2.14 m

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2020 (09 Jan Shift 2)
A wire of length L and radius r is clamped at one end. If its other end is pulled by a force F, its length increases by l. If the radius of the wire and the applied force both are reduced to half of their original values keeping original length constant, the increase in length will become:

Options:
A 3 times
B 32 times
C 4 times
D 2 times

The correct answer is:
2 times

Solution:
The formula to calculate the Young's modulus of the material of the wire, in the first case, can be written as

Y=Fπr2lL ...1

From equation (1), it follows that

F=Yπr2×lL ...2

For the second case, the Young's modulus is given by

Y=F/2πr2/4l'L ...3

From equation (3), it follows that

F=Yl'L×2×πr24 ...4

From equations (2) and (4), it implies that

Yπr2×lL=Yl'L×2×πr24⇒l'=2l

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2024 (29 Jan Shift 2)
A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force F, its length increases by 5 cm. Another wire of the same material of length 4 L and radius 4r is pulled by a force 4 F under same conditions. The increase in length of this wire is _____ cm.

Options:

The correct answer is:
5

Solution:
Young's modulus is given by Y=FLA∆L, here, A is cross-section area and ∆L is change in length.

Now, when force F is applied, change in length ΔL1=FLAY=FLπr2Y=5 cm

Now, when force 4F is applied, change in length

ΔL2=4 F4Lπ16r2Y=FLπr2Y=5 cm

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2022 (25 Jul Shift 1)
A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force f, its length increases by l. Another wire of same material of length 2L and radius 2r is pulled by a force 2f. Then the increase in its length will be:

Options:
A 4l
B l2
C 2l
D l

The correct answer is:
l

Solution:
Young's modulus is same for both the wires since the material is same.

For the first wire,

Y=fLπr2l ...(i)

For the second wire,

Y=2f(2L)4πr2l' ...(ii)

Dividing (i) by (ii)

fLπr2l×4πr2l'4fL=1⇒l'=l

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2023 (15 Apr Shift 1)
A wire of length L carrying a current I is bent into a circle. The magnitude of the magnetic field at the centre of the circle is

Options:
A πμ0IL
B μ0I2L
C 2πμ0IL
D μ0I2πL

The correct answer is:
πμ0IL

Solution:
Length L= circumference of circle =2πR
⇒ R=L2π
∴ Magnetic field at the centre of circular current loop is
B=μ0I2R=μ0I2 L×2π
B=μ0πIL

Topic: Physics, Magnetic Effects of Current, JEE Main
A wire of length l has a resistance R. If half of the length is stretched to make the radius half of its original value, then the final resistance of the wire is

Options:
A 9R
B 5R
C 17R2
D 3R

The correct answer is:
17R2

Solution:
R= ρl A = ρl π r 2
π r 2 l 2 =π ( r 2 ) 2 l ′
l ′ =2l
R'=ρl'A'=ρ2lπr22=8R
R''=ρl2πr2=R2

Req=R'+R''=17R2

Topic: Physics, Current Electricity, JEE Main
A wire of length $l$ is bent into a circular loop of radius $R$ and carries a current $I$. The magnetic field at the centre of the loop is $B$. The same wire is now bent into a double loop of equal radii. If both loops carry the same current $I$ and it is in the same direction, the magnetic field at the centre of the double loop will be

Options:
A Zero
B 2 B
C 4 B
D 8 B

The correct answer is:
Zero

Solution:
Magnetic field at the centre of the loop
For the wire which is looped double let radius becomes $r$ Then,
$\begin{aligned}
\frac{l}{2} & =2 \pi r \\
\text{or} \quad \frac{l}{4 \pi} & =(r)
\end{aligned}$
$\begin{array}{ll}\therefore & B^{\prime}=\frac{\mu_0}{4 \pi} \cdot \frac{I \cdot 2 \pi r \times 2}{r^2} \\ \text { or } & B^{\prime}=\frac{\mu_0}{4 \pi} \cdot \frac{I \cdot \frac{l}{2} \cdot 2}{\left(\frac{l}{4 \pi}\right)^2}\end{array}$
Dividing Eq. (ii) by Eq. (iii), we get
$\frac{B^{\prime}}{B}=\frac{\frac{\mu_0}{4 \pi} \cdot \frac{I \cdot l \cdot 16 \pi^2}{l^2}}{\frac{\mu_0}{4 \pi} \cdot \frac{I l \cdot 4 \pi^2}{l^2}}$
$\begin{array}{ll}\text { or } & \frac{B^{\prime}}{B}=4 \\ \text { or } & B^{\prime}=4 B\end{array}$

Topic: Physics, Magnetic Effects of Current, JEE Main
A wire of length l tapers uniformly from end P to end Q with the diameter at P twice that at Q. A potential difference is applied across the ends of the wire. Which graph represents the variation of the drift velocity V of the conduction electrons with distance x?

Options:
A
B
C
D

The correct answer is:

Solution:
For a current-carrying conductor
neÏr2V=i=constantâ´V=KÏr2â´ V â 1r2
Now at x=0 radius=2r0 ; at x =l radius=r0
â´Vat x=0=KÏ4r02Vat x=l=KÏr02

Topic: Physics, Current Electricity, JEE Main
A wire of resistance $5 \Omega$ is drawn out so that its new length is 3 times its original length. What is the reistance of the new wire?

Options:
A $45 \Omega$
B $15 \Omega$
C $5 / 3 \Omega$
D $5 \Omega$

The correct answer is:
$45 \Omega$

Solution:
$\left(\frac{r_1}{r_2}\right)^2=\left(\frac{\ell_2}{\ell_1}\right)=\frac{3 \ell}{\ell}=3$
$\left(\frac{R_2}{R_1}\right)=\frac{\ell_2}{\ell_1} \times \frac{A_1}{A_2}=3 \times\left(\frac{r_1}{r_2}\right)^2=3 \times 3 \Rightarrow R_2=45$

Topic: Physics, Current Electricity, JEE Main
A wire of resistance $12 \mathrm{ohms}$ per metre is bent to form a complete circle of radius $10 \mathrm{~cm}$. The resistance between its two diametrically opposite points, A and B as shown in the figure, is :

Options:
A $6 \Omega$
B $0.6 \pi \Omega$
C $3 \Omega$
D $6 \pi \Omega$

The correct answer is:
$0.6 \pi \Omega$

Solution:
Total resistance of wire $=12 \Omega \times 2 \pi \times 10^{-1}$
$$
=2.4 \pi
$$
Resistance of each half $=\frac{2.4 \pi}{2}=1.2 \pi$
and as about diameter both parts are in parallel
$$
\text { Req. }=\frac{1.2 \pi}{2}=0.6 \pi \Omega
$$

Topic: Physics, Current Electricity, NEET, NEET 2009 (Mains)
A wire of resistance 160 Ω is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be

Options:
A 16 Ω
B 10 Ω
C 640 Ω
D 40 Ω

The correct answer is:
10 Ω

Solution:
The formula to calculate the resistance of the original wire is given by

R=ρlA ...1

When the new wire is formed, its resistance is given by

R'=ρl'4A ...2

Divide equation (2) by equation (1) and simplify to obtain the expression for the new resistance.

R'R=ρl'4AρlA⇒R'=Rl'4l ...3

Substitute the values of the known parameters into equation (3) to calculate the required new resistance.

R'=l44l×160 Ω= 10 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (12 Apr Shift 1)
A wire of resistance R1$R_{1}$ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:

Options:
A 9:1
B 1:9
C 4:1
D 3:1

The correct answer is:
9:1

Solution:
As per the given question, the change in the length is ∆l=2l.

Resistance of a wire is given by, R=ρlA

Therefore, R1=ρl1 A1

When the length is starched by 2l1, final length will be l2=l1+2l1=3l1. As the volume remains constant l2A2=l1A1 ⇒A2=A13. Therefore,

R2=ρ3l1 A13=9ρl1 A1

∴R2R1=9

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (28 Jul Shift 1)
A wire of resistance R and length L is cut into 5 equal parts. If these parts are joined parallely, then resultant resistance will be :

Options:
A R25
B R5
C 25R
D 5R

The correct answer is:
R25

Solution:
Resistance of each part will be equally divided. So, the resistance of each part of the wire can be expressed as R5.

The formula to calculate the equivalent resistance of the combination of resistors is given by

1Req=∑i=151Ri ...1

From equation (1), it follows that

1Req=5R+5R+5R+5R+5R=25R⇒Req=R25

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (27 Jan Shift 1)
A wire of resistance $R$ and radius $r$ is stretched till its radius became $r / 2$. If new resistance of the stretched wire is $x R$, then value of $x$ is _______

Options:

The correct answer is:
16

Solution:
We know $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}, \mathrm{R} \propto \frac{l}{\mathrm{r}^2}$
As we starch the wire, its length will increase but its radius will decrease keeping the volume constant
$\begin{aligned}
& \mathrm{V}_{\mathrm{i}}=\mathrm{V}_{\mathrm{f}} \\
& \pi \mathrm{r}^2 l=\pi \frac{\mathrm{r}^2}{4} l_{\mathrm{f}} \\
& l_{\mathrm{f}}=4 l \\
& \frac{\mathrm{R}_{\text {new }}}{\mathrm{R}_{\text {old }}}=\left(\frac{4 l}{\frac{\mathrm{r}^2}{4}}\right) \frac{\mathrm{r}^2}{l}=16 \\
& \mathrm{R}_{\text {new }}=16 \mathrm{R} \\
\therefore & \mathrm{x}=16
\end{aligned}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (06 Apr Shift 1)
A wire of Young's modulus $1 \cdot 6 \times 10^{12} \mathrm{~N} / \mathrm{m}^{2}$ is stretched by a force so as to produce a strain of $2 \times 10^{-4}$. The energy density of the wire is

Options:
A $3 \cdot 2 \times 10^{4} \mathrm{~J} / \mathrm{m}^{3}$
B $3 \cdot 2 \times 10^{8} \mathrm{~J} / \mathrm{m}^{3}$
C $1 \cdot 6 \times 10^{3} \mathrm{~J} / \mathrm{m}^{3}$
D $6 \cdot 4 \times 10^{3} \mathrm{~J} / \mathrm{m}^{3}$

The correct answer is:
$3 \cdot 2 \times 10^{4} \mathrm{~J} / \mathrm{m}^{3}$

Solution:
Energy $=1 / 2 \times Y \times(\text { Strain })^{2}$
$=1 / 2 \times 1.6 \times 10^{12} \times 4^{2} \times 10^{-8}$
$=3.2 \times 10^{4}$.

Topic: Physics, Mechanical Properties of Solids, JEE Main
A wire, suspended vertically from one of its ends, is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm. The elastic energy (in J) stored in the wire is

Options:

The correct answer is:
0.1

Solution:
U=12ÃFÃl=12Ã200Ã10-3=0.1 J

Topic: Physics, Work Power Energy, JEE Main
A wire suspended vertically from one of its ends is stretched by attaching a weight of $200 \mathrm{~N}$ to the lower end. The weight stretches the wire by $1 \mathrm{~mm}$. Then the elastic energy stored in the wire is

Options:
A
$0.2 \mathrm{~J}$

B
$10 \mathrm{~J}$

C
$20 \mathrm{~J}$

D
$0.1 \mathrm{~J}$

The correct answer is:
$0.1 \mathrm{~J}$

Solution:
Elastic energy $=\frac{1}{2} \times \mathrm{F} \times \mathrm{x}$
$$
\mathrm{F}=200 \mathrm{~N}, \mathrm{x}=1 \mathrm{~mm}=10^{-3} \mathrm{~m} \quad \therefore \mathrm{E}=\frac{1}{2} \times 200 \times 1 \times 10^{-3}=0.1 \mathrm{~J}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2003
A wire suspended vertically from one of its ends is stretched by attaching a weight of $200 \mathrm{~N}$ to the lower end. The weight stretches the wire by $1 \mathrm{~mm}$. Then the elastic energy stored in the wire is

Options:
A $0.2 \mathrm{~J}$
B $10 \mathrm{~J}$
C $20 \mathrm{~J}$
D $0.1 \mathrm{~J}$

The correct answer is:
$0.1 \mathrm{~J}$

Solution:
Elastic energy per unit volume $=1 / 2 \times$ stress $\times$ strain
Elastic Energy $=1 / 2 x$ stress $x$ strain $x$ volume
$\begin{aligned} & =1 / 2 \times F / A \times(\Delta \mathrm{L} / \mathrm{L}) \times(\mathrm{AL}) \\ & =1 / 2 \times F \Delta \mathrm{L} \\ & =1 / 2 \times 200 \times 10^{-3}\end{aligned}$
Elastic Energy $=0.1 \mathrm{~J}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm. Then the elastic energy stored in the wire is

Options:
A 0.2 J
B 10 J
C 20 J
D 0.1 J

The correct answer is:
0.1 J

Solution:
Elastic energy per unit volume = 1 2 × stress × strain
∴ Elastic energy = 1 2 × stress × strain × volume
= 12×FA×ΔLL×AL
=12FΔL=12×200×10-3=0.1 J

Topic: Physics, Mechanical Properties of Solids, JEE Main
A wire when connected to $220 \mathrm{~V}$ mains supply has power dissipation $\mathrm{P}_1$. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is $P_2$ Then $P_2: P_1$ is

Options:
A
1

B
4

C
2

D
3

The correct answer is:
4

Solution:
$P_1=V^2 / R ; P_2=\frac{V^2}{(R / 2)}+\frac{V^2}{(R / 2)}=4 \frac{V^2}{R}=4 P_1$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2002
A wire $X$ is half the diameter and half the length of a wire $Y$ of similar material. The ratio of resistance of $\mathrm{X}$ to that of $\mathrm{Y}$ is

Options:
A 8: 1
B 4: 1
C 2: 1
D 1: 1

The correct answer is:
2: 1

Solution:
$\mathrm{R} \propto \frac{\ell}{\mathrm{D}^{2}} \Rightarrow \frac{\mathrm{Rx}}{\mathrm{Ry}}=\frac{2}{1}$

Topic: Physics, Current Electricity, JEE Main
A wire X of length 50 cm carrying a current of 2 A is placed parallel to a long wire Y of length 5 m. The wire Y carries a current of 3 A. The distance between two wires is 5 cm and currents flow in the same direction. The force acting on the wire Y is :

âââââââ

Options:
A 1.2×10-5 N directed towards wire X.
B 1.2×10-4 N directed away from wire X.
C 1.2×10-4 N directed towards wire X.
D 2.4×10-5 N directed towards wire X.

The correct answer is:
1.2×10-5 N directed towards wire X.

Solution:
An attractive force will act normal to X towards Y because the direction of the currents in the wires is the same.

Force exerted on length l due to the magnetic field is given as F=FXY=FYX=I1l1 B12

⇒F=μ0I1I22πrl1

Given here, I1=2 A, I2= 3 A, l1=5 m, r=5 cm.

Putting the values, we have

F=4π×10-7×2×32π×5×10-2×50×10-2

=1.2×10-5 N directed towards wire X.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (29 Jul Shift 2)
A : With change in concentration of the reactant, the value of equilibrium constant changes. \( R \) : More the value of \( K_{C} \) lesser is the extent of reaction in the forward direction.

Options:
A If both Assertion & Reason are true and the reason is the correct explanation of the assertion
B If both Assertion & Reason are true but the reason is not the correct explanation of the assertion
C If Assertion is true statement but Reason is false
D If both Assertion and Reason are false statements

The correct answer is:
If both Assertion and Reason are false statements

Solution:
Value of equilibrium constant does not depend on concentration of reactants or products.

Topic: Chemistry, Chemical Equilibrium, JEE Main
A wooden block floating in a bucket of water has 45 of its volume submerged. When certain amount of an oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is:

Options:
A 0.8
B 0.7
C 0.5
D 0.6

The correct answer is:
0.6

Solution:
Case-1, σ Vg=ρw45Vg …….(I)
Case-2, σ Vg=ρwV2g+ρoV2g …….(II)
From eq (I) & (II),
45ρw=ρw2+ρo2
45-12ρw=ρo2
8-510ρw=ρo2
ρoρw=35=0.6

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2019 (09 Apr Shift 2)
A wooden block is floating on water kept in a beaker. $40 \%$ of the block is above the water surface. Now the beaker is kept inside a lift that starts going upward with acceleration equal to $g / 2$. The block will then

Options:
A $\operatorname{sink}$
B float with $10 \%$ above the water surface
C fioat with $40 \%$ above the water surface
D float with $70 \%$ above the water surface

The correct answer is:
fioat with $40 \%$ above the water surface

Solution:
The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it, if the fluid is in rest. In this case, the fluid is accelerating upwards so the buoyant force must also provide the displaced fluid force to accelerate.
Therefore, buoyant force will be \(\mathrm{fV} \rho \mathrm{g}_{\text {eff }}\) where \(\mathrm{V}=\) volume of body and \(\mathrm{f}=\) fraction of volume of body immersed in fluid and \(\mathrm{g}_{\text {eff }}=\mathrm{g}+\mathrm{a}=1.5 \mathrm{~g}\).
When fluid is at rest: \(\mathrm{f}=1-0.4=0.6\),
so \(0.6 \mathrm{~V} \rho \mathrm{g}=\mathrm{V} \rho_{\mathrm{b}} \mathrm{g}\)
where \(\rho_{\mathrm{b}}\) is the density of the block. \(\Rightarrow \rho_{\mathrm{b}}=0.6 \rho .\)
In the second case: \(1.5 \mathrm{fV} \rho \mathrm{g}=\mathrm{V} \rho_{\mathrm{b}} \mathrm{g}+\mathrm{V} \rho_{\mathrm{b}} \mathrm{a}=1.5 \mathrm{~V} \rho_{\mathrm{b}} \mathrm{g} \Rightarrow \mathrm{f} \rho=\rho_{\mathrm{b}} \Rightarrow \mathrm{f}=0.6\) Thus, the fraction of immersed volume remains the same.
Body will float with \(40 \%\) of the volume above water surface.

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2014
A wooden block of mass 1 kg and density 800 kg m-3 is held stationary, with the help of a string, in a container filled with water of density 1000 kg m-3, as shown in the figure. If the container is moved upwards with an acceleration of 2 m s-2, then the tension in the string will be ( take g=10 m s-2 )

Options:
A 2 N
B 3 N
C 5 N
D 6 N

The correct answer is:
3 N

Solution:
Wâ²=m(g+a)=1Ã12=12 N
Bâ²=ÏlÃvÃ(g+a)=1000Ã1800Ã(12)=15 N
â´ T=Bâ²âWâ²=15â12=3 N

Topic: Physics, Mechanical Properties of Fluids, JEE Main