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QUESTION: 1

If weight of an object at pole is 196 N then weight at equator is [g = 10 m/s^{2} ; radius of earth = 6400 Km]

Solution:

at pole, weight = mg = 196

m = 19.6 kg

at equator, weight = mg – mω^{2}R

= 195.33 N

QUESTION: 2

In a house 15 Bulbs of 45 W, 15 bulbs of 100 W, 15 bulbs of 10 W and Two heaters of 1 KW each is connected to 220 V Mains supply then find minimum fuse current

Solution:

Total power is (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000)

= 4325 W

So current is = 4325/220

= 19.66 A

Ans is 20 Amp.

QUESTION: 3

In an adiabatic process, volume is doubled then find the ratio of final average relaxation time & initial relaxation time. Given

Solution:

relaxation time

and

QUESTION: 4

A block of mass 10kg is suspended from string of length 4m. When pulled by a force F along horizontal from midpoint. Upper half of string makes 45° with vertical, value of F is

Solution:

QUESTION: 5

The surface mass density of a disc of radius a varies with radial distance as σ = A + Br where A & B are positive constants then moment of inertia of the disc about an axis passing through its centre and perpendicular to the plane

Solution:

σ = A + Br

I =

= (A + Br )2πr^{3} dr

= 2pa^{4}

QUESTION: 6

Cascaded Carnot engine is an arrangement in which heat sink of one engine is source for other. If high temperature for one engine is T_{1}, low temperature for other engine is T_{2} (Assume work done by both engine is same) Calculate lower temperature of first engine.

Solution:

Let, Q_{H} : Heat input to I^{st} engine

Q_{L} : Heat rejected from I^{st}

engine

Q_{L}': Heat rejected f rom IInd engine

Work done by I^{st }engine = work done by II^{nd} engine

Q_{H} – Q_{L} = Q_{L} – Q_{L}'

2Q_{L} = Q_{H} + Q_{L}'

QUESTION: 7

Activity of a substance changes from 700 s^{–1} to 500 s^{–1} in 30 minute. Find its half-life in minutes

Solution:

=> ln2 = λt_{1/2}...........(i)

(i)/(ii)

(2.06004) 30 = t_{1/2} = 61.8 min.

QUESTION: 8

In YDSE, separation between slits is 0.15 mm, distance between slits and screen is 1.5 m and wavelength of light is 589 nm, then fringe width is

Solution:

= 5.9mm

QUESTION: 9

An ideal fluid is flowing in a pipe in streamline flow. Pipe has maximum and minimum diameter of 6.4 cm and 4.8 cm respectively. Find out the ratio of minimum to maximum velocity.

Solution:

Using equation of continuity

A_{1}V_{1} = A_{2}V_{2}

QUESTION: 10

There is a electric circuit as shown in the figure. Find potential difference between points a and b.

Solution:

Diode is in forward bias, so it will behave as simple wire so,

So V_{ab} = = 10V

QUESTION: 11

A particle of mass m and positive charge q is projected with a speed of v_{0} in y–direction in the presence of electric and magnetic field are in x–direction. Find the instant of time at which the speed of particle becomes double the initial speed.

Solution:

As(magnitude of velocity does not change in y–z plane)

∴

QUESTION: 12

Two sources of sound moving with same speed v and emitting frequency of 1400 Hz are moving such that one source s_{1} is moving towards the observer and s_{2} is moving away from observer. If observer hears beat frequency of 2 Hz. Then find the speed of source. Given

V_{sound} >> V_{Source}

V_{sound} = 350 m/s

Solution:

QUESTION: 13

An electron & a photon have same energy E. Find the ratio of de Broglie wavelength of electron to wavelength of photon. Given mass of electron is m & speed of light is C.

Solution:

λ_{d} for electron =

λ for photon = hC/E

Ratio =

QUESTION: 14

A ring is rotated about diametric axis in a uniform magnetic field perpendicular to the plane of the ring. If initially the plane of the ring is perpendicular to the magnetic field. Find the instant of time at which EMF will be maximum & minimum respectively :

Solution:

When ωt = π

∵ φ will have maximum.

∵ e will be minimum.

t = = 5 sec.

QUESTION: 15

Electric field in space is given by A positively charged particle at (0, 0, π/K) is given velocity v_{0}kˆ at t = 0. Direction of f orce acting on particle is

Solution:

Force due to electric field is in direction

because at t = 0, E =

Force due to magnetic field is in direction

∴ it is parallel to

∴ net force is antiparallel to

QUESTION: 16

Focal length of convex lens in air is 16 cm (μ_{glass} = 1.5). Now the lens is submerged in liquid of refractive index 1.42. Find the ratio of focal length in medium to focal length in air has closest value

Solution:

QUESTION: 17

A lift of mass 920 kg has a capacity of 10 persons. If average mass of person is 68 kg. Friction f orce between lift and lift shaft is 6000 N. The minimum power of motor required to move the lift upward with constant velocity 3 m/s is [g = 10 m/s^{2}]

Solution:

Net force on motor will be

Fm = [920 + 68(10)]g + 6000

= 22000 N

So, required power for motor

= 22000 x 3

= 66000 watt

QUESTION: 18

The hysteresis curve for a material is shown in the figure. Then for the material retentivity, coercivity and saturation magnetization respectively will be

Solution:

x = retentivity

y = coercivity

z = saturation magnetization

QUESTION: 19

An inductor of inductance 10 mH and a resistance of 5Ω is connected to a battery of 20 V at t = 0. Find the ratio of current in circuit at t = ∞ to current at t = 40 sec.

Solution:

QUESTION: 20

Find the dimension of

Solution:

Energy density in magnetic field =

*Answer can only contain numeric values

QUESTION: 21

A capacitor of 60 pF charged to 20 volt. Now battery is removed and then this capacitor is connected to another identical uncharged capacitor. Find heat loss in nJ.

Solution:

V_{0} = 20 V

Heat loss = Ui – Uf

= 6 × 10^{–9 }J = 6 nJ

*Answer can only contain numeric values

QUESTION: 22

When m gram of steam at 100°C is mixed with 200 gm of ice at 0°C. it results in water at 40°C. Find the value of m in gram .

(given : Latent heat of fusion (L_{f}) = 80 cal/gm, Latent heat of vaporisation (L_{v}) = 540 cal/gm., specific heat of water (C_{w})= 1 cal/gm/°C)

Solution:

*Answer can only contain numeric values

QUESTION: 23

A solid cube of side 'a' is shown in the figure. Find maximum value of for which the block does

not topple before sliding.

Solution:

For no toppling

For no toppling

0.2a + 0.4b ≤ 0.5a

0.4b ≤ 0.3a

b ≤ 0.75a (in limiting case)

But it is not possible as b can maximum be equal to 0.5a

∴

*Answer can only contain numeric values

QUESTION: 24

Magnitude of resultant of two vectors is equal to magnitude of . Find the angle between

and resultant of 2and

Solution:

So angle between (2 + ) and is 90°

Alternate solution

P^{2} + Q^{2} + 2PQcosθ = P^{2}

Q + 2Pcosθ =θ

cosθ =

tan α = = ∞

∵ [2Pcosθ + Q = 0]

α = 90°

*Answer can only contain numeric values

QUESTION: 25

A battery of unknown emf connected to a potentiometer has balancing length 560 cm. If a resistor of resistance 10W is connected in parallel with the cell the balancing length change by 60 cm. If the internal resistance of the cell is n/10 Ω, the value of 'n' is

Solution:

Let the emf of cell is e internal resistance is 'r' and potential gradient is x.

only cell connected :

ε = 560 x ................. (i)

After connecting the resistor

................... (ii)

from (1) and (2)

56 = 50 + 5r

n = 12

QUESTION: 26

Which of the following reactions are possible?

(A)

(B)

(C)

(D)

Solution:

Vinyl halides and aryl halides do not give Friedel craft's reaction.

QUESTION: 27

A and B are in the given reaction ?

Solution:

QUESTION: 28

The correct statement about gluconic acid is

Solution:

Gluconic acid

is obtained by partial oxidation of glucose by Tollen's reagent or Fehling solution or Br_{2},H_{2}O.

Gluconic acid can not form hemiacetal or acetal

QUESTION: 29

Stability order of following alkoxide ions is

Solution:

When negative charge is delocalised with electron withdrawing group like (NO_{2}) then stability increases.

(A) Negative charge is delocalised with NO_{2} group

(B) Negative charge is delocalised with carbon of alkene

(C) Negative charge is localised

QUESTION: 30

A and B are –

Solution:

QUESTION: 31

For the complex [Ma_{2}b_{2}] if M is sp^{3} or dsp^{2} hybridised respectiv ely then total number of optical isomers are respectively :

Solution:

Both will not show optical isomerism.

QUESTION: 32

Bond order and magnetic nature of CN^{–} are respectively

Solution:

CN^{–} is a 14 electron system.

QUESTION: 33

Which of the following is incorrect?

Solution:

Theory based.

QUESTION: 34

NaOH + Cl_{2} → A + other products

**Hot & conc**.

Ca(OH) + Cl_{2} → B + other products

**Cold & dil.**

A & B are respectively

Solution:

6NaOH + 3Cl_{2} → 5NaCl + NaClO_{3} + 3H_{2}O

2Ca(OH)_{2} + Cl_{2} → Ca(OCl)_{2} + CaCl_{2} + H_{2}O

QUESTION: 35

There are two beakers (I) having pure volatile solvent and (II) having volatile solvent and non-volatile solute. If both beakers are placed together in a closed container then:

Solution:

There will be lowering in vapour pressure in second beaker.

QUESTION: 36

Metal with low melting point containing impurities of high melting point can be purified by

Solution:

Theory based

QUESTION: 37

Which of the following statements are correct ?

(I) On decomposition of H_{2}O_{2}, O_{2} gas is released .

(II) 2-ethylanthraquinol is used in preparation of H_{2}O_{2}

(III) On heating KClO_{3}, Pb(NO_{3})_{2}, NaNO_{3}, O_{2} gas is released.

(IV) In the preparation of sodium peroxoborate, H_{2}O_{2} is treated with sodium metaborate.

Solution:

Theory based

QUESTION: 38

Amongs the following which is redox reaction?

Solution:

N_{2} + O_{2} → 2NO

3O_{2} → 2O_{3 }

2NaOH + H_{2}SO_{4} → Na_{2}SO_{4} + 2H_{2}O

AgNO_{3} + NaCl → NaNO_{3} + AgCl

QUESTION: 39

Select the correct options :

Solution:

Option a is correct. The graph is for no of molecules vs speed. So don’t go for values in Vrms,Vavg and Vmp. It's a derivation according to Maxwell velocity distribution of gas. Most gas particles have the most probable velocity. Some of the particles have avg velocity while few particles have root mean square velocity.

QUESTION: 40

Which one of the following amongs each pair will release maximum energy on gaining one electron (A = F, Cl), (B = S, Se), (C = Li, Na)

Solution:

Theory based

QUESTION: 41

Which of the following statements are incorrect ?

(A) Co^{+3} with strong field ligand forms high magnetic moment complex.

(B) For Co^{+3} if pairing energy(P) > Δ_{o} then the complex formed will have t^{4}_{2g}, e^{2}_{g} configuration

(C) For [Co(en)_{3}]^{3+ }λ_{absorbed} is less than λ_{absorbed} for [CoF_{6}]^{3-}

(D) If Δ_{o} = 18000 cm^{-1} for Co^{+3} then with same ligands for it Δ_{t} = 16000 cm^{-1}

Solution:

Theory based

QUESTION: 42

0.6 g of urea on strong heating with NaOH evolves NH_{3}. Liberated NH_{3} will combine completely with which of the following HCl solution ?

Solution:

2 × mole of Urea = mole of NH_{3}...(1)

mole of NH_{3 }= mole of HCl ........(2)

∴ mole of HCl = 0.02 mole

*Answer can only contain numeric values

QUESTION: 43

Number of sp^{2} hybrid carbon atoms in aspartame is –

Solution:

All stared carbon atoms of aspartame are sp2 hybrid. Aspartame is methyl ester of dipeptide formed from aspartic acid and phenylalanine.

*Answer can only contain numeric values

QUESTION: 44

3 gram of acetic acid is mixed in 250 mL of 0.1 M HCl. This mixture is now diluted to 500 mL. 20 mL of this solution is now taken is another container mL of 5M NaOH is added to this. Find the pH of this solution. (log 3 = 0.4771, pKa = 4.74)

Solution:

m mole of acidic acid in 20 mL = 2

m mole of HCl in 20 mL = 1

m mole of NaOH = 2.5

pH = PKa + log

= 4.74 + log 3

= 4.74 + 0.48 = 5.22

*Answer can only contain numeric values

QUESTION: 45

Flocculation value for As_{2}S_{3} sol by HCl is 30 m mole L^{–1}. Calcualte mass of H_{2}SO_{4} required in gram for 250 mL sol.

Solution:

For 1L sol 30 m mol of HCl is required

∴ For 1L sol 15 m mol H_{2}SO_{4} is required

For 250 mL of sol

m mol H_{2}SO_{4} = 0.3675 g

*Answer can only contain numeric values

QUESTION: 46

Determine total number of atoms in per unit formula of (A), (B) & (C)

Solution:

(A) = CrO_{2}Cl_{2}

(B) = Na_{2}CrO_{4}

(C) = CrO_{5}

*Answer can only contain numeric values

QUESTION: 47

Calculate Δ_{f}H° (In kJ/mol) for C_{2}H_{6}(g), if Δ_{c}H° [C_{(graphite)}] = -393.5 kJ/mol,

Δ_{C}Hº [H_{2}(g)] = -286 kJ/mol and

ΔCHº [C_{6}H_{6}(g)] = -1560 kJ/mol

Solution:

QUESTION: 48

Let A = [aij], B=[bij] are two 3 × 3 matrices such that bij = λ^{i + j - 2}aij & |B| = 81. Find |A| if λ = 3.

Solution:

=>81 = 3^{3}.3. 3^{2} |A|

=> 3^{4 }= 3^{6} |A|

=> |A|= 1/9

QUESTION: 49

From any point P on the line x = 2y perpendicular is drawn on y = x. Let foot of perpendicular is Q. Find the locus of mid point of PQ.

Solution:

slope of PQ = = -1

⇒ k – α = - h + 2α

⇒ α = ......... (1)

Also 2h = 2α + β

2k = α + β

⇒ 2h = α + 2k

⇒ a = 2h - 2k .....(2)

from (1) & (2)

= 2(h - k)

so locus is 6x –6y = x + y

⇒ 5x = 7y

QUESTION: 50

Pair of tangents are drawn from origin to the circle x^{2} + y^{2} – 8x – 4y + 16 = 0 then square of length of chord of contact is

Solution:

L = √S_{1} = √16 = 4

R = = 2

Length of Chord of contact =

square of length of chord of contact = 64/5

QUESTION: 51

Contrapositive of if A ⊂ B and B ⊂ C then C ⊂ D

Solution:

Let P = A ⊂ B, Q = B ⊂ C, R = C ⊂ A

Contrapositive of (P ∧ Q) → R is ~ R → ~ (P ∧ Q)

R ν (~ P ν ~ Q)

QUESTION: 52

Let y(x) is solution of differential equation (y^{2} – x) dx/dy = 1 and y(0) = 1, then find the value of x where curve cuts the x-axis

Solution:

c = –e

y = 0, x = 0 – 0 + 2 + (–e) (e^{–0})

x = 2 –e

QUESTION: 53

Let θ_{1} and θ_{2} (whereθ_{1} < θ_{2})are two solutions of 2cot^{2}θ θ ∈ [0, 2π) then is equal to

Solution:

2 cos^{2}θ - 5sinθ + 4sin2θ = 0, sinθ ≠0

QUESTION: 54

Let 3 + 4 + 8 + 9 + 13 + 14 + 18 +……….40 terms = S. If S = (102)m then m =

Solution:

S = 7 + 17 + 27 + 37 + 47 +......20 terms

= 10[2040] = (102) (20)

⇒ m = 20

QUESTION: 55

If 6 , then number of ordered pairs (r, k) are –(where k ∈ I).

Solution:

Hence number of order pair = 4

QUESTION: 56

Let 4α = 5 then a =

Solution:

⇒ 4(2 - e^{-α}a - e-^{2a}) = 5

Put e^{-α} = t

⇒ 4t^{2} + 4t – 3 = 0

⇒(2t + 3) (2t - 1) = 0

⇒

⇒ a = ln2

QUESTION: 57

Let f(x) is a five degree polynomial which has critical points x = ±1 and = 4 then which one is incorrect.

Solution:

f(x) = ax^{5 } +bx^{4} + cx^{3}

= 4

⇒ 2 + c = 4

⇒ c = 2

f'(x) = 5ax^{4} + 4bx^{3} + 6x^{2}

=

f'(1) = 0 ⇒ 5a + 4b + 6 = 0

f'(–1) = 0 ⇒ 5a – 4b + 6 = 0

b = 0,

f'(x) = –6x^{4} + 6x^{2}

= 6x^{2} (–x^{2} +1)

= –6x^{2} (x+1) (x–1)

Minimal at x = –1

Maxima at x = 1

x = –1

x = 1

QUESTION: 58

If are unit vectors such that and and then

Solution:

⇒ λ = -3/2

QUESTION: 59

Coefficient of x^{7 }in is

Solution:

Coefficient of x^{7} is ^{11}C_{7 }= ^{11}C_{4} = 330

QUESTION: 60

Let α and β are the roots of x^{2} - x - 1 = 0 such that P_{k} = α^{k} + β^{k}, k ≥ 1 then which one is incorrect?

Solution:

= 5(1) + 6

P_{5} = 11 and P_{5} = α^{2} + β^{2} = α + 1 + β + 1

P_{2} = 3 and

P_{3} = α^{3} + β^{3} = 2α + 1 + 2β + 1

= 2(1) +2 = 4

P_{2} × P_{3} = 12 and P_{5} = 11

⇒ P^{5} ≠ P_{2} × P_{3}

QUESTION: 61

Let f(x) = x^{3} –4x^{2 }+ 8x + 11, if LMVT is applicable on f(x) in [0, 1], value of c is :

Solution:

f(x) is a polynomial function

∴ it is continuous and differentiable in [0, 1]

Here f(0) = 11, f(1) = 1 – 4 + 8 + 11 = 16 f '(x) = 3x^{2} – 8x + 8

∴

= 3c^{2} – 8c + 8

⇒ 3c^{2} – 8c + 3 = 0

∴

QUESTION: 62

The area bounded by 4x^{2 }≤ y ≤ 8x + 12 is -

Solution:

4x^{2} = y

y = 8x + 12

4x^{2} = 8x + 12

x^{2} – x – 3 = 0

x^{2} – 2x – 3 = 0]

x^{2} – 3x + x – 3 = 0

(x + 1)(x – 3) = 0

x = –1

= 128/3

QUESTION: 63

There are 5 machines. Probability of a machine being faulted is 1/4 Probability of atmost two machines s faulted, is k then value of k is

Solution:

Required probability = when no. machine has fault + when only one machine has fault + when only two machines have fault.

∴ k = 17/8

QUESTION: 64

3x + 4y = 12√2 is the tangent to the ellipse then the distance between focii of ellipse is-

Solution:

condition of tangency c^{2} = a^{2}m^{2} + b^{2}

a^{2} = 16

1a = 4

∴ focus are (± 7 , 0)

∴ distance between foci = 2√7

QUESTION: 65

If z = is purely real and θ = then arg(sinθ + I cosθ) is -

Solution:

as z is purely real

⇒ 3 cosθ + 4sinθ = 0

⇒ tanθ =

arg(sinθ + icosθ) =

QUESTION: 66

a_{1}, a_{2}, a_{3} …..a_{9} are in GP where a_{1} < 0,

a_{1 }+ a_{2} = 4, a_{3} + a_{4} = 16, if = 4λ, then λ is equal to

Solution:

a_{1} + a_{2} = 4

⇒ a_{1} + a_{1r} = 4 ................... (i)

⇒ r^{2 }= 4

r = ±2

r = 2, a_{1}(1+2) = 4

⇒ a_{1}= 4/3

r = –2, a_{1}(1–2) = 4

⇒ a_{1} = - 4

λ = –171

QUESTION: 67

If and .Then dy/dx at x = 1/2 is

Solution:

*Answer can only contain numeric values

QUESTION: 68

Let X = {x : 1 ≤ x ≤ 50, x ∈ N}

A = {x: x is multiple of 2}

B = {x: x is multiple of 7}

Then find number of elements in the smallest subset of X which contain elements of both A and B

Solution:

*Answer can only contain numeric values

QUESTION: 69

If Qis foot of perpendicular drawn from P(1, 0, 3) on a line L and if line L is passing through (α, 7, 1), then value of α is

Solution:

Since PQ is perpendicular to L, therefore

*Answer can only contain numeric values

QUESTION: 70

If f(x) is defined in x ∈

f(x) =

Find k such that f(x) is continuous

Solution:

= 3 + 2 = 5

∴ f(x) will be continuous if f(0) =

*Answer can only contain numeric values

QUESTION: 71

If system of equation

x + y + z = 6

x + 2y + 3z = 10

3x + 2y + λz = μ has more than two solutions. Find (μ – λ^{2})

Solution:

x + y + z = 6 …….. (1)

x + 2y + 3z = 10 …….. (2)

3x + 2y + λz = μ ..... (3)

from (1) and (2)

if z = 0

⇒ x + y = 6 and x + 2y = 10

⇒ y = 4, x = 2

(2, 4, 0)

if y = 0 ⇒ x + z = 6 and x + 3z = 10

⇒ z = 2 and x = 4

(4, 0, 2)

so, 3x + 2y + λz = μ must pass through (2, 4, 0) and (4, 0, 2)

so, 6 + 8 = μ

⇒ μ= 14 and 12 + 2λ = μ

12 + 2λ = 14

⇒ λ = 1

so μ – λ^{2} = 14 - 1 = 13

*Answer can only contain numeric values

QUESTION: 72

If mean and variance of 2, 3, 16, 20, 13, 7, x, y are 10 and 25 respectively then find xy

Solution:

mean =

⇒ x + y = 19.............(i)

variance = 25

⇒ x^{2} + y^{2} = 113………(ii)

(x+y)^{2} = (19)^{2}

⇒ x^{2} + y^{2} + 2xy = 361 ⇒ xy = 124 (exact data is not retrieved so ans. can vary)

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