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JEE Main Practice Test- 15 - JEE MCQ


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75 Questions MCQ Test - JEE Main Practice Test- 15

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JEE Main Practice Test- 15 - Question 1

The equation to the directrix of a parabola if the two extremities of its latus rectum are (2, 4) and (6, 4) and the parabola passes through the point (8, 1) is

Detailed Solution for JEE Main Practice Test- 15 - Question 1

focus is (4, 4) & D can be y = 6 or y = 2

where ‘O’ is origin and S is the focus and D is directrix

JEE Main Practice Test- 15 - Question 2


then the maximum value of  Δ is

Detailed Solution for JEE Main Practice Test- 15 - Question 2

Apply R3 → R3 → R1, we get

= (3cos θ – sin θ)2 So, maximum value of Δ equals 10.

JEE Main Practice Test- 15 - Question 3

The number of solution(s) of the equation 


(where z = x + iy, x, y ∈ R, i2 = –1 and x ≠ 2)

Detailed Solution for JEE Main Practice Test- 15 - Question 3



⇒ x = – √2
∴ z = – √2
Hence only one z will satisfy above equation.

JEE Main Practice Test- 15 - Question 4

Two circles of radii r1 and r2 are both touching the coordinate axes and intersecting each other orthogonally. The value of r1/r2 (where r1 > r2) equals

Detailed Solution for JEE Main Practice Test- 15 - Question 4

Circle is (x – r)2 + (y – r)2 = r2
⇒ x2 + y2 – 2xr – 2yr + r2 = 0
Hence the circles are x2 + y2 – 2xr1 – 2yr1 + r12 = 0    ......(1)

x2 + y2 – 2xr2 – 2yr2 + r22 = 0 .....(2)
As (1) and (2) are orthogonal so
2r1r2 + 2r1r2 = r12 + r22


JEE Main Practice Test- 15 - Question 5

Let X and Y be two matrices satisfying this relations

then Tr.(X) – Tr.(Y) equals
[Note : Tr.(P) denotes trace of matrix P.]

JEE Main Practice Test- 15 - Question 6

The differential equation dx/dy = 3y/2x represents a family of hyperbolas (except when it represents a pair of lines) with eccentricity can be

Detailed Solution for JEE Main Practice Test- 15 - Question 6

2x dx – 3y dy = 0 gives, on integration, 
 
The solution represents a family of hyperbolas given by

whose eccentricity


and eccentricity

it gives a pair of lines which are the asymptotes of the hyperbolas.

JEE Main Practice Test- 15 - Question 7

Line L, perpendicular to the line with equation y = 3x – 5, contains the point (1, 4). The x-intercept of L, is

Detailed Solution for JEE Main Practice Test- 15 - Question 7

put y = 0, x = 13

JEE Main Practice Test- 15 - Question 8

Let A = [aij] (1 ≤ i , j ≤ 3) be a 3 × 3 matrix and B = [bij] (1 ≤ i , j ≤ 3) be a 3 × 3 matrix such that


If det. A = 4, then the value of det. B is

Detailed Solution for JEE Main Practice Test- 15 - Question 8

B = AAT.
Hence, det.
B = |AAT| = |A| |AT| = |A|2 = 42 = 16.

JEE Main Practice Test- 15 - Question 9

Number of words that can be formed using all the letters of the word GARGEE if no two alike letters are together, is

Detailed Solution for JEE Main Practice Test- 15 - Question 9

Total – n(A ∪ B)


Set A represents number of ways when G's are together
Set B represents number of ways when E's are together

Aliter: GG EE A R
Number of words when

Number of words when G's are separated but E's are together = 3! × 4C2 = 36
∴ Number of ways when no two alike letters are together = 120 – 36 = 84

JEE Main Practice Test- 15 - Question 10

If acute angle between the line  and xy plane is α and acute angle between the planes x + 2y = 0 and 2x + y = 0 is β then (cos2α + sin2β) equals

Detailed Solution for JEE Main Practice Test- 15 - Question 10

JEE Main Practice Test- 15 - Question 11

If area of pentagon PQRST be 7, where P(–1, –1), Q(2, 0), R(3, 1), S(2, 2) and T(–1, t), t > 0, then the value of t is

Detailed Solution for JEE Main Practice Test- 15 - Question 11


Area of pentagon PQRST = 7
⇒ ar.(trapezium PQST) + ar.(ΔQRS) = 7

⇒ t = 1

JEE Main Practice Test- 15 - Question 12

The sum of all value of λ for which the lines 2x + y + 1 = 0; 3x + 2λy + 4 = 0; x + y - 3λ = 0 are concurrent, is

Detailed Solution for JEE Main Practice Test- 15 - Question 12


(3λ + 1)(2λ + 1) + 3λ(2λ - 4) = 0
⇒ 6λ2 + 5λ + 1 + 6λ2 - 12λ = 0
⇒ 12λ2 - 7λ + 1 = 0
⇒ (3λ – 1)(4λ – 1) = 0

⇒ Sum = 7/12 

JEE Main Practice Test- 15 - Question 13

If A and B are two independent events such that P(A' ∩ B') = 2/15, P(A ∩ B') = 1/6 then P(B) =

JEE Main Practice Test- 15 - Question 14

A hyperbola has centre at origin and one focus at (6, 8). If its two directrices are 3x + 4y + 10 = 0 and 3x + 4y –10 = 0 and eccentricity is e,then the value of 4e2/5 is equal to

Detailed Solution for JEE Main Practice Test- 15 - Question 14

Distance between centre and focus = ae = 10
Distance between directrices = 2a/e = 4

JEE Main Practice Test- 15 - Question 15

Number of integral values of 'k' for which the chord of the circle x2 + y2 = 125 passing through P(8, k) gets bisected at P (8, k) and has integral slope is

Detailed Solution for JEE Main Practice Test- 15 - Question 15

The slope of the chord is 

⇒ k = ± 1, ± 2, ± 4, ± 8 but (8, k) must also lie inside the circle x2 + y2 = 125

⇒ 64 + k2 – 125 < 0
⇒ k2 < 61
⇒ k can be equal to ± 1, ± 2, ± 4
⇒ 6 values

JEE Main Practice Test- 15 - Question 16

Locus of the feet of perpendiculars drawn from points (1, 2) and (3, 4) on a variable tangent to the conic | z - (1+ 2i) | - | z - (3+ 4i) | = 2 is

Detailed Solution for JEE Main Practice Test- 15 - Question 16


represents a hyperbola with foci (1, 2) and (3, 4) and length of transverse axis = 2.
∴ 2a = 2 ⇒ a = 1
∵ Feet of perpendiculars from foci on any tangent lie on auxilliary circle of the hyperbola.
∴ Locus will be auxilliary circle.
∴ Centre = mid point of foci = (2, 3)

and radius = semi transverse axis = 1
∴ Equation of auxilliary circle is |z - (2 + 3i) |= 1

JEE Main Practice Test- 15 - Question 17

Number of numbers greater than a million and divisible by 5 which can be formed by using only the digits 1, 2,1, 2, 0, 5 and 2 is :

Detailed Solution for JEE Main Practice Test- 15 - Question 17

JEE Main Practice Test- 15 - Question 18

The dual of statement (p v ~ q) ∧ (~ p) is

Detailed Solution for JEE Main Practice Test- 15 - Question 18

In dual statement v replace by ∧ and ∧ replace by v so answer is (p ∧ ~ q) v (~ p).

JEE Main Practice Test- 15 - Question 19

A normal is drawn to the parabola y2 = 9x at the point P(4, 6), S being the focus, a circle is described on the focal distance of the point P as diameter. The length of the intercept made by the circle on the normal at P is

Detailed Solution for JEE Main Practice Test- 15 - Question 19

Required intercept will be equal to the perpendicular distance from the focus on the tangent at P.
Tangent at P,


⇒ 12y = 9x + 36
⇒ 9x - 12y + 36 = 0

JEE Main Practice Test- 15 - Question 20

Consider ellipse  Let C is centre of the ellipse and P is a variable point lying on the ellipse. If the angle between CP and tangent at P is minimum, then P may be

Detailed Solution for JEE Main Practice Test- 15 - Question 20


and C (0, 0)


∴ angle is minimum, when θ = 45°

*Answer can only contain numeric values
JEE Main Practice Test- 15 - Question 21

(Instruction to attempt numerical value (integer) type question: If your answer is 100 write 100 only. Do not write 100.0)

If then the value of a5 + b4 must be


Detailed Solution for JEE Main Practice Test- 15 - Question 21

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JEE Main Practice Test- 15 - Question 22

The value of x satisfying the equation

then k is


Detailed Solution for JEE Main Practice Test- 15 - Question 22

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JEE Main Practice Test- 15 - Question 23

The sum of squares of all integral values of a for which the quadratic expression (x−a)(x−10)+1 can be factored as a product (x+α)(x+β) of two factors and α, β ∈ I must be equal to


Detailed Solution for JEE Main Practice Test- 15 - Question 23

*Answer can only contain numeric values
JEE Main Practice Test- 15 - Question 24

If f(x) = tan-1 (sin x + cos x)3 is an increasing function, then the value of x in (0, 2π) is x ∈  Then the value of a + 10b + 100c + 1000d must be


Detailed Solution for JEE Main Practice Test- 15 - Question 24

*Answer can only contain numeric values
JEE Main Practice Test- 15 - Question 25

The sum of two digit even numbers which do not end with zero is


Detailed Solution for JEE Main Practice Test- 15 - Question 25

Required sum
= (12 + 14 + 16 + …. + 98) – (20 + 30 + 40 + …. + 90)
44/2 (12 + 98) – 8/2 (20 + 90)
= 22(110) – 4(110)
= (110) (18)
= 1980

JEE Main Practice Test- 15 - Question 26

A non-conducting rod AB of length l has a total charge q. The rod is rotated about an axis passing through its center of mass with a constant angular velocity ω as shown in the figure. The magnetic moment of the rod is

Detailed Solution for JEE Main Practice Test- 15 - Question 26

JEE Main Practice Test- 15 - Question 27

The distance between two parallel plates of a capacitor is a. A conductor of thickness b(b < a) is inserted between the plates as shown in the figure. The variation of effective capacitance between the plates of the capacitor as a function of the distance (x) is best represented by

Detailed Solution for JEE Main Practice Test- 15 - Question 27

JEE Main Practice Test- 15 - Question 28

A solid sphere of radius R, and dielectric constant ‘k’ has spherical cavity of radius R/4. A point charge q1 is placed in the cavity. Another charge q2 is placed outside the sphere at a distance of r from q1. Then Coulombic force of interaction between them is found to be ‘F1’. When the same charges are separated by same distance in vacuum then the force of interaction between them is found to be F2 then

Detailed Solution for JEE Main Practice Test- 15 - Question 28

Coulombic force between them remains same.

Charge flown from battery = CV
Work done = CV2
Heat produced ΔH = ΔU + ΔW

JEE Main Practice Test- 15 - Question 29

Energy stored in the capacitor in it’s steady state is

Detailed Solution for JEE Main Practice Test- 15 - Question 29

Potential across capacitor is zero, hence energy stored is zero.

JEE Main Practice Test- 15 - Question 30

A point charge of 0.1C is placed on the circumference of a non-conducting ring of radius 1m which is rotating about an axis passing from centre and perpendicular to the plane of ring with a constant angular acceleration of 1 rad/sec2. If ring starts from rest at t = 0, the magnetic field at the centre of the ring at t = 10 sec, is

Detailed Solution for JEE Main Practice Test- 15 - Question 30

ω = 0 + 1 × 10 = 10 rad/sec2
∴ v = rω = 1 × 10 = 10 m/s

JEE Main Practice Test- 15 - Question 31

In an L – C circuit shown in the figure, C = 1F, L = 4H. At time t = 0, charge in the capacitor is 4C and it is decreasing at a rate of √5 C/s. Choose the correct statements.

Detailed Solution for JEE Main Practice Test- 15 - Question 31

JEE Main Practice Test- 15 - Question 32

A solid conducting sphere of radius r is having a charge Q and point charges +q and –q are kept at distances d from the center of sphere as shown in the figure. The electric potential at the centre of solid sphere

Detailed Solution for JEE Main Practice Test- 15 - Question 32

JEE Main Practice Test- 15 - Question 33

Consider the circuit in the adjacent figure. What will be potential difference between A and B in the steady state

Detailed Solution for JEE Main Practice Test- 15 - Question 33

There will be no current any where in the circuit.

JEE Main Practice Test- 15 - Question 34

A charge q is placed at some distance along the axis of a uniformly charged disc of surface charge density σ C/m2. The flux due to the charge q through the disc is ϕ. The electric force on charge q exerted by the disc is

Detailed Solution for JEE Main Practice Test- 15 - Question 34


JEE Main Practice Test- 15 - Question 35

In the given circuit diagram, find the heat generated on closing the switch S. (Initially the capacitor of capacitance C is uncharged)

Detailed Solution for JEE Main Practice Test- 15 - Question 35

Only charge is that capacitor 'C' will get charged.

JEE Main Practice Test- 15 - Question 36

A metallic ring of radius R moves in a vertical plane in the presence of a uniform magnetic field B perpendicular to the plane of the ring. At any given instant of time its centre of mass moves with a velocity v while ring rotates in its COM frame with angular velocity ω as shown in the figure. The magnitude of induced e.m.f. between points O and P is

Detailed Solution for JEE Main Practice Test- 15 - Question 36


JEE Main Practice Test- 15 - Question 37

An object is placed at 30 cm from a convex lens of focal length 15cm. On the other side of the lens a convex mirror of focal length 12cm is placed so that the principal axis of both coincide. It is observed that the object and image coincide (autocollimation). What is the separation between the lens and mirror?

Detailed Solution for JEE Main Practice Test- 15 - Question 37

For image to be coincident, either the rays should retrace or the image due to the lens should formed just at the pole of the mirror in thin case. The image formed due to lens is at 30 cm (2f) be from the lens. Thus either this image should be at centre of curvature of the convex mirroror at the pole of the mirror. Hence 6cm or 30cm should be the separation between the lens and the mirror.

JEE Main Practice Test- 15 - Question 38

The relation between R and r (internal resistance of the battery) for which the power consumed in the external part of the circuit is maximum

Detailed Solution for JEE Main Practice Test- 15 - Question 38

Its a wheat stone bridge with equivalent 2R.

JEE Main Practice Test- 15 - Question 39

A capacitor and resistor are connected with an A.C. source as shown in figure. Reactance of capacitor is XC = 3W and resistance of resistor is 4Ω. Phase difference between current I and I1 is

Detailed Solution for JEE Main Practice Test- 15 - Question 39

Let I2 be current in capacitor


JEE Main Practice Test- 15 - Question 40

Find the stress at distance R/2 from centre in a uniformly charged non conducting sphere having radius R and charge density ρ.

Detailed Solution for JEE Main Practice Test- 15 - Question 40




JEE Main Practice Test- 15 - Question 41

The capacitor is initially uncharged. Find ratio of current through the 10 Ω resistance and through the 20 Ω resistance initially.

Detailed Solution for JEE Main Practice Test- 15 - Question 41


JEE Main Practice Test- 15 - Question 42

In the diagram below, light is incident on the interface between media 1 and 2 as shown and is totally reflected. The light is then also totally reflected at the interface between media 1 and 3, after which it travels in a direction opposite to its initial direction. The two interfaces are perpendicular. The refractive indices are related as

Detailed Solution for JEE Main Practice Test- 15 - Question 42





∴ from (1) and (2), n12 – n22 > n32

JEE Main Practice Test- 15 - Question 43

Which of the following transitions of He+ ion will give rise to spectral line which has same wavelength as some spectral line in hydrogen atom ?

Detailed Solution for JEE Main Practice Test- 15 - Question 43

If n2 → n1 in H (z = 1) gives λ then z n2 → z n1 gives λ in H-like ion for He+ ion, z = 2

JEE Main Practice Test- 15 - Question 44

Two imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of the concentric spheres. When compared to the number of field lines N1 going through the sphere of radius R, the number of electric field lines N2 going through the sphere of radius 2R is

Detailed Solution for JEE Main Practice Test- 15 - Question 44

JEE Main Practice Test- 15 - Question 45

A radioactive sample contains two radioactive nucleus A and B having decay constant λ hr–1 and 2λ hr–1. Initially 20% of decay comes from A. How long (in hr) will it take before 50% of decay comes from A. [Take λ = ln 2]

Detailed Solution for JEE Main Practice Test- 15 - Question 45



∴ from (1) and (2) , eλt = 9
⇒ λt = 2ln3 ⇒ t = 2.

*Answer can only contain numeric values
JEE Main Practice Test- 15 - Question 46

(Instruction to attempt numerical value (integer) type question: If your answer is 100 write 100 only. Do not write 100.0)

A block is placed on an inclined plane moving towards right horizontally with an acceleration a0=g. The length of the plane AC = 1m. Friction is absent everywhere. Find the time taken (in seconds) by the block to reach from C to A.


Detailed Solution for JEE Main Practice Test- 15 - Question 46

*Answer can only contain numeric values
JEE Main Practice Test- 15 - Question 47

In the given figure, find the horizontal velocity u (in ms-1) of a projectile so that it hits the inclined plane perpendicularly. Given H = 6.25 m:


Detailed Solution for JEE Main Practice Test- 15 - Question 47

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JEE Main Practice Test- 15 - Question 48

A uniform cylinder rests on a cart as shown. The coefficient of static friction between the cylinder and the cart is 0.5. If the cylinder is 4 cm in diameter and 10 cm in height, then what is the minimum acceleration (in m/s2) of the cart needed to cause the cylinder to tip over?


Detailed Solution for JEE Main Practice Test- 15 - Question 48

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JEE Main Practice Test- 15 - Question 49

In the circuit shown in figure, find the ratio of currents i1/i2


Detailed Solution for JEE Main Practice Test- 15 - Question 49

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JEE Main Practice Test- 15 - Question 50

The centers of two identical small conducting spheres are 1m apart. They carry charges of opposite kind and attract each other with a force F. When they are connected by a conducting thin wire, they repel each other with a force F/3. The ratio of the magnitude of charges carried by the spheres initially is n : 1. Find the value of n.


Detailed Solution for JEE Main Practice Test- 15 - Question 50

JEE Main Practice Test- 15 - Question 51

A 0.1 M solution of which salt is most acidic?

JEE Main Practice Test- 15 - Question 52

Which is the strongest acid amongst the following compounds?

Detailed Solution for JEE Main Practice Test- 15 - Question 52

JEE Main Practice Test- 15 - Question 53

Select the correct order of metallic character :

Detailed Solution for JEE Main Practice Test- 15 - Question 53

As we move left to right metallic character decreases and as we move top to bottom metallic character increases, so correct is

JEE Main Practice Test- 15 - Question 54

In hydrogen atom which transition produces a photon with highest energy?

JEE Main Practice Test- 15 - Question 55

The correct basicity order of indicated atoms P, Q and R is -

Detailed Solution for JEE Main Practice Test- 15 - Question 55

Basicity order of indicated atoms P, Q, R is P > Q > R

JEE Main Practice Test- 15 - Question 56

Select the incorrect statement regarding N3–, O2–, F-, Na+ and Mg2+.

JEE Main Practice Test- 15 - Question 57

Dioxygen difluoride (O2F2) is a highly oxidising and unstable liquid. At 300 K it decomposes back to oxygen and fluorine, which are both gases at this temperature. The equation for the reaction is given below. 0.1 g of O2F2 was left for 24 hours and the 24.9 ml of gas mixture evolved was collected at 300 K and 100 kPa. What % by mass of dioxygen difluride has decomposed by this time?
O2F2(l) → O2(g) + F2(g)

Detailed Solution for JEE Main Practice Test- 15 - Question 57


JEE Main Practice Test- 15 - Question 58

Which of the following cyclic dienes does not show geometrical isomerism?

Detailed Solution for JEE Main Practice Test- 15 - Question 58


Cyclohexene does not show
GeometricalIsomerism

JEE Main Practice Test- 15 - Question 59

Which of the following property is different for two different isoelectronic homonuclear diatomic species?

JEE Main Practice Test- 15 - Question 60

In an analysis of solutions containing Barium ions (Ba2+), 50 ml of solution gave 0.233 g of BaSO4 upon addition of sufficient sulphuric acid to precipitate all the Ba2+ ions present. What is concentration (in M) of Ba2+ ions in the solution.

Detailed Solution for JEE Main Practice Test- 15 - Question 60

JEE Main Practice Test- 15 - Question 61

Ethers are more volatile than same no. of carbon containing alcohol due to -

Detailed Solution for JEE Main Practice Test- 15 - Question 61

Ethers are more volatile than same number of carbon containing alcohol due to absence of H-bonding.

JEE Main Practice Test- 15 - Question 62

Select the species which becomes Bent due to lone pair – bond pair repulsion.

Detailed Solution for JEE Main Practice Test- 15 - Question 62




(Bent, due to lone pair-bond pair repulsion)

JEE Main Practice Test- 15 - Question 63

Quinaldine red is a useful acid-base indicator which is red in solution of pH greater than 3.5 but colorless below pH = 1.5. Which of the following solution would turn red if a few drops of quinaldine red were added?
(i) 0.1 mol L–1 HCl
(ii) 0.05 mol L–1 NH3
(iii) 0.0001 mol L–1 CH3COOH

JEE Main Practice Test- 15 - Question 64


Identify P and Q respectively.

Detailed Solution for JEE Main Practice Test- 15 - Question 64

JEE Main Practice Test- 15 - Question 65

Oxidising nature of H2 is observed on reaction with

Detailed Solution for JEE Main Practice Test- 15 - Question 65



(3) Cu2O + H2 → 2Cu + H2O
→ act as reducing agent
(4) RCHO + H2 → R – CH2 – OH
→ act as reducing agent

JEE Main Practice Test- 15 - Question 66

A brown-black compound of thallium was found to contain 89.5% Tl and 10.5% oxygen. What is oxidation number of thallium in this compound? [Atomic weight of Tl = 204]

Detailed Solution for JEE Main Practice Test- 15 - Question 66

Empirical formula of the compound = Tl

JEE Main Practice Test- 15 - Question 67

Identify R in the following series of reaction.

Detailed Solution for JEE Main Practice Test- 15 - Question 67

JEE Main Practice Test- 15 - Question 68

Select the incorrect match regarding action of H2O.

Detailed Solution for JEE Main Practice Test- 15 - Question 68


JEE Main Practice Test- 15 - Question 69

Identify the correct statement :

JEE Main Practice Test- 15 - Question 70

In the mono chlorination of 3-Ethyl pentane, how many total products will be formed?

Detailed Solution for JEE Main Practice Test- 15 - Question 70


Total products = 4

*Answer can only contain numeric values
JEE Main Practice Test- 15 - Question 71

(Instruction to attempt numerical value (integer) type question: If your answer is 100 write 100 only. Do not write 100.0)

From the following sets of quantum numbers, state which are not possible.
a. n = 0, l = 0, m = 0, s = +1/2
b. n = 1, l = 0, m = 0, s = –1/2
c. n = 1, l = 1, m = 0, s = +1/2
d. n = 1, l = 0, m = +1, s = +1/2
e. n = 3, l = 0, m = –1, s = –1/2
f. n = 2, l = 2, m = 0, s = –1/2
g. n = 2, l = 1, m = 0, s = –1/2


Detailed Solution for JEE Main Practice Test- 15 - Question 71

a. is not possible because n = 0 is wrong.
b. is possible.
c. is not possible because l = n is wrong.
d. is not possible because m = + 1 is wrong.
e. is not possible
f. is not possible because l = 2 is wrong.
g. is possible

*Answer can only contain numeric values
JEE Main Practice Test- 15 - Question 72

200g impure CaCO3 on heating gives 5.6L. CO2 gas at STP. Find the percentage of Calcium in the limestone sample.


Detailed Solution for JEE Main Practice Test- 15 - Question 72

CaCO3 → CaO + CO2
5.6/ 22.4 = 0.25 mole
Mole of CaO = mole of Ca = 0.25
Mass of Ca= 0.25 x 40 = 10
% of Calcium = 10 x 100/ 200 = 5%

*Answer can only contain numeric values
JEE Main Practice Test- 15 - Question 73

NO → N2O + NO2
n this reaction equivalent weight of NO is x/y M then x + y will be


Detailed Solution for JEE Main Practice Test- 15 - Question 73

NO → N2O (-2 to + 1 for one atom ) = 3
NO → NO(+2 to + 4) = 2

*Answer can only contain numeric values
JEE Main Practice Test- 15 - Question 74

10 g impure NaOH is completely neutralised by 1000 ml of 1/10 N HCI. Calculate the percentage purity of the impure NaOH.


Detailed Solution for JEE Main Practice Test- 15 - Question 74

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JEE Main Practice Test- 15 - Question 75


Then weight of MgSO4 will be formed in this series is


Detailed Solution for JEE Main Practice Test- 15 - Question 75

MgCO3 Molar mass: 84.3139
Moles = 0.05 moles produced 0.05 moles MgO and
MgSO4 molar mass: 120 x 0.05 = 6

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