BRILLIANT/DYANPRASAD ACADEMY, PANDHARPUR

STD – XII Physics CET TEST-4

1 1
1. Ans:- C = 5 × 103−10 = 5 × 10−7
1 10 × 106
∴ = = 2 × 106 /m
λ 5

6. Ans:- C
7. Ans:- B
In half revolution, the point A of the wheel
reaches at position A’, so displacements is,
AA′ = √(AB)2 + (A′ B)2

= √(πR)2 + (2𝑅)2 = 𝑅√𝜋 2 + 4

= 1√𝜋 2 + 4 = √𝜋 2 + 4 (∵ 𝑅 = 1𝑚)
2. Ans:- B If c is the velocity of light in vacuum, then
c
Percentage error in R.I. of the medium = v
m
2∆𝐵 1 ∆𝑎 ∆𝑐
X=( +2 𝑥 + ) c 3 × 108
𝑏 𝑎 𝑐 ∴ vm = n = n
1
= ( 2 x 3% + x 4% + 2% ) 1 1 1
2 But R. I. (n) = sin 𝐶 = sin 300 = 1/2 = 2
= ( 6 % + 2% + 2 % )
3 × 108
= 10 % ∴ vm = = 1.5 × 108 m/s
2

8. Ans:- (D)
1 2 q q
3. Ans:-D F = 4πϵ(r/2)2

⃗𝑨
⃗ 𝑿 ⃗𝑩
⃗ = q q 4q1 q2
F ′ = 4πϵ(r/2)
1 2
2 = 4πϵr2
𝑖̂ 𝑗̂ 𝑘̂
𝐹′ 𝜖0 1 1
1 3 −1 =4 = 4 (𝐾) = 4 (4) = 1
𝐹 𝜖
2 −1 1
′
∴ F =F
⃗𝑨
⃗ 𝑿 ⃗𝑩
⃗ = ( 3-1 ) 𝑖̂ − ( 1-2 ) 𝑗̂ + (−1 − 6 )𝑘
9. Ans:- B
⃗⃗ = 𝟐𝑖̂ - 3𝑗̂-7𝑘̂
⃗⃗ 𝑿 𝑩
𝑨
λ 6000 Å 3
an m = λ1 = 4000 Å = 2
⃗ l = √4 + 9 + 49 = √62
𝑙 𝐴𝑋𝐵 2

2 2
⃗ =
Unit vector parallel to 𝐴 𝑋 𝐵 ∴ mn a = 3 = sin C ∴ C = sin−1 (3)
𝐴̂ 𝑥 𝐵̂ ̂
2𝑖̂−3𝑗̂ −7𝑘
̂=
𝒏 = 10. Ans:- B
𝑙 𝐴𝑋𝐵 ⃗ l √62
Let u and θ be the velocity and angle of
4. Ans:- C
projection of a projectile. Maximum height
5. Ans:- C
1 1
attained
Wave number = = 5000 × 10−10
𝜆

Prof. Mandar V. Paricharak M.Sc., B.Ed, M.Ed Page (1) Prof. Vinayak D. Paricharak M.Sc., B.Ed
 u2 sin2 θ 25 2
H= = 20 × ( 5 ) = 20 × 25 = 500𝑁
2g

Horizontal range, 15. Ans:-D

u2 2 sin θ cos θ 𝑑𝑙 (4−2)
R= 𝑒 = 𝐿 𝑑𝑡 𝑜𝑟 8 = 𝐿
g 0.05

R 8 ×0.05
= 4 cot θ = C(= a constant) ∴ 𝐿= = 0.2 𝐻
H 2

∆R ∆H 16. Ans:- (B)

=
R H
𝑅2 𝑇 4 32 ×105 𝑇 4
% increase in horizontal range = (𝑇2 ) ⟹ 2
= (400)
𝑅1 1 2 ×105
∆R ∆H 4
= × 100 = × 100 = 6% 2
∴ 16 = (400)
𝑇
R H

11. Ans:- (A) ∴ 𝑇2 = 2 × 400 = 800𝐾 ⟹ 5270C

v = √μrg 17. Ans:- (B)
= √0.4 × 50 × 9.8 = √196 = 14𝑚/𝑠 𝜏 = 𝐼𝛼 ; 𝑖. 𝑒. 500 = 𝐼 × 25
12. Ans:- (A) 𝐼 = 20 𝑘𝑔 𝑚2 = 𝑀𝐾 2
20
The graph of K.E. against frequency is a 𝐾 2 = 50 = 0.4 𝑖. 𝑒. 𝐾 ≈ 0.6 𝑚
straight line. The slope of the graph is ‘h’ i.e.
18. Ans:- (C)
the Planck’s constant. At it is a universal
𝑊 = 𝑇 × 𝑑𝐴
constant, it does not depend upon the
= 5 × (2 × 0.02) = 5 × 0.04 = 2 × 10−1 J
intensity of radiation and has the same value
19. Ans:- (B)
for all metals.
Max K. E. = hv − W0 = 6.2 − 4.2 = 2 eV
= 2 × 1.6 × 10−19 = 3.2 × 10−19 J
Type equation here.
20. Ans:- (D)
In cyclic process, net work done = Area of
cycle = Area ACBDA
or
Net work done during the complete cycle is
13. Ans:- (D) W = Area ACBB’A’A – Area ADBB’A’A
1 1 W = Area ACBDA
𝑋𝐶 = 𝜔𝐶 = 2𝜋𝑣𝐶
1 21. Ans:- (B)
= 2𝜋(50) ×1/𝜋 = 10−2 Ω
R = √a21 + a22 + 2a1 a2 cos(α1 − α2 )
14. Ans:- (C)
𝜋 𝜋
𝐹1
= 𝐴2
𝐹 = √16 + 9 + 2(4)(3)𝑐𝑜𝑠 ( 6 − 3 )
𝐴1 2

∴ 𝐹2 = 𝐹1 × 𝐴2
𝐴 = √25 + 24 × 0.8660 = √25 + 20.784
1
= √45.784 = 6.766 𝑐𝑚

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22. Ans:-A deflection is 10 mA
𝜇 𝐵 1.6
𝜇𝑟 = 𝜇 = 𝜇 = 4𝜋 × 10−7 × ∴ High resistance R is given by
0 0𝐻 1000
𝑉 25
= 1.3 × 103 𝑅 = 𝐼 − 𝐺 = 10 ×10−3 − 20
𝑔

𝜇𝑟 = 1 + 𝜒 ∴ 𝑇 = 2500 − 20 = 2480 Ω
∴ 𝜒 = 𝜇𝑟 − 1 We can also use R = G (n – 1) where n = 125
= 1300 − 1 = 1299. 26. Ans:- (D)
23. Ans:- (A) 𝜔2 = 2
For a circular arc, subtending an angle θ at its 4𝜋 2
=2
𝑇2
centre,
4π2
μ θ. i ∴ T2 =
B = 4π0 [ ] 2
r 1
3𝜋 ∴ 𝑇 = √2𝜋 𝑖. 𝑒. 𝑛 = per second
For a circle, θ = 2π. In this case, 𝜃 = √2𝜋
2
27. Ans :- D
𝜇 3𝜋 𝑖
∴ 𝐵 = 4𝜋0 [ 2 × 𝑟]
3 𝜇0 𝑖
∴ 𝐵 = 8. 28. Ans:- (B)
𝑟
mv2
24. Ans:- (B) T= + mg cos θ = 0.85 N
r
The revolving electron is equivalent to 29. Ans:- (B)
𝑞
current 𝐼 = 𝐶0 𝑇 𝑀𝐻𝑒
𝑡
= √𝑇 0 ×
𝐶𝐻𝑒 𝐻𝑒 𝑀0
But for the electron, q = e and
2𝜋𝑟 𝑇 𝑀𝐻𝑒
𝑡=𝑇= 1 = √𝑇 0 ×
𝑣 𝐻𝑒 𝑀0
e ev
∴ I= 2πr = 2πr ∴ 𝑇0 =
𝑇𝐻𝑒 × 𝑀0
=
273 × 32
= 2184𝐾
v 𝑀𝐻𝑒 4
𝜇 2𝜋𝐼
the field at the centre of the coil = 4𝜋0 30. Ans:- (A)
𝑟
𝐼 𝑒𝑣 𝑣 Temperature is a measurement of coldness or
or 𝐹 ∝ 𝑟 𝑖. 𝑒. 𝐹 ∝ 2𝜋𝑟 × 𝑟 𝑜𝑟 𝐹 ∝ 𝑟 2
hotness of an object. The definition is based
e
(∵ is a constant) on Zeroth law of thermodynamics.
2π
1
But in Bohr’s model v ∝ n 31. Ans:- (D)
Pi 4000
Where n is the quantum number of the orbit As Ep Ip = Pi ∴ IP = = = 40 A
Ep 100

and 𝑟 ∝ 𝑛2 32. Ans:- C

𝑣
But 𝐹 ∝ 𝑟 2 Energy needed, E = 2 LI2
1

1
∴ 𝐹 ∝ 𝑛 (𝑛2 )2 However, no energy is required to maintain
1 the magnetic flus in the coil.
∴ 𝐹 ∝ 𝑛5
33. Ans:- A
25. Ans:- (B)
The current for 50 div i.e. for full scale
Prof. Mandar V. Paricharak M.Sc., B.Ed, M.Ed Page (3) Prof. Vinayak D. Paricharak M.Sc., B.Ed
 μ I dl sin θ
34. Ans:- (D) element at a point is dB = 4π0 r2
When a bright fringe is formed opposite to
For straight sections, angle between ⃗⃗⃗⃗
𝑑𝐼 𝑎𝑛𝑑 𝑟
𝑑
one of the slits, 𝑥 = 2 is 00 or π. Therefore, magnetic field at point
xd d d d2
Path difference = = 2 × D = 2D O due to straight sections is zero. Magnetic
D

If it is nth order bright fringe, path difference, field at point O due to semicircular are is
1 𝜇0 𝐼 1 4𝜋 × 10−7 ×12
d2
nλ = 2D or n = 2Dλ
d2 𝐵 =2× =2× = 1.89 × 10−4 𝑇
2𝑟 2 ×2 × 10−2

35. Ans:- (C) 42. Ans:- C

ϕ 2 ×10−2
At Brewster’s angle, only the reflected light is M= = = 2H
I 0.01
plane polarised, but transmitted light is 43. Ans:- C
partially polarised. When current in the circuit is wattles, power
36. Ans:-B consumed in the circuit is zero. Now P =
V V 3000
E= or d = E = =6m Ev Iv cos ϕ or 0 = Ev Iv cos ϕ.
d 500

37. Ans:- A Now Ev and Iv are not zero so that

𝑑𝑉
𝐸𝑥 = − 𝑑𝑥 = −[−2𝑥𝑦 − 𝑧 3 ] cos 𝜙 = 0 𝑖. 𝑒. 𝜙 = 900 .

𝑑𝑉
44. Ans:- (A)
𝐸𝑦 = − 𝑑𝑦 = −[−𝑥 2 ] hC
E = mC2 and E = hv =
𝑑𝑉 λ
𝐸𝑧 = − 𝑑𝑦 = −[−3𝑥𝑧 2 ] ℎ𝐶 h h
∴ = 𝑚𝐶 2 ∴ λ = mC = p
𝜆
⃗⃗⃗ = 𝐸𝑥 𝑖̂ + 𝐸𝑦 𝐽̂ + 𝐸𝑧 𝑘̂
𝐸
45. Ans:- (A)
∴ 𝐸⃗ = 𝑖̂(2𝑥𝑦 + 𝑧 3 ) + 𝑗̂ + 𝑥 2 + 𝑘̂3 × 𝑧 2 𝐸 2 1
𝐼=𝑅 = 1000 = 500 𝐴
1 + 𝑅2
38. Ans:- (B)
10 1
𝑦 = 3 cos 2𝜋 (50𝑡 − 𝑥/2) ∴ P. D. across the wire = IR = 500 = 50 V
V 1
∴ 𝜆 = 2 𝑐𝑚 ∴ Potential gradient = = 50 ×2 = 0.01 V/m
L
39. Ans:- (B) 46. Ans:- B
40. Ans:- (D) When the two capacitors are connected in
The activity cannot be increased or decreased parallel, there is redistribution of charge till
by any method. Radioactivity is a the two capacitiors acquire a common
spontaneous phenomenon. It is independent potential V given by:
of the external physical conditions like 𝐶1 𝑉1 + 𝐶2 𝑉2
𝑉= 𝐶1 + 𝐶2
temperature, pressure, etc.
(20 × 10−6 × 500) + (10 × 10−6 × 200)
41. Ans:-A =
(20 × 10−6 ) + (10 × 10−6 )
The wire is divided into three sections; two
= 400 V
straight sections and one semicircular arc.
47. Ans:- (B)
The magnetic field due to current carrying
Here, a = 2 mm = 2 × 10−3 𝑚
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 𝜆 = 500 𝑛𝑚 = 500 × 10−9
= 5 × 10−7 𝑚, 𝐷 = 1 𝑚
The distance between the first minima on
2𝜆𝐷 2 × 5 × 10−7 × 1
either side on a screen = =
𝑎 2 × 10−3

= 5 × 10−4 m = 0.5 × 10−3 m = 0.5mm

48. Ans:- D

m1 m2 Fr2
F=G ∴ G=m
r2 1 m2

Unit of G – Nm2/kg2 and Dimensional

equation of G is
Fr2 M1 L1 T−2 × L2
[G] = [ ]=[ ]
m1 m2 M2

∴ [𝐺] = [𝑀−1 𝐿3 𝑇 −2 ]
49. Ans:- C
50. Ans:- (C)
B = 𝜇0 𝑛𝐼
𝐼 3
B’ = 𝜇0 (3𝑛 ) × = 2 [𝜇0 𝑛𝐼 ]
2
3
= B
2

Prof. Mandar V. Paricharak M.Sc., B.Ed, M.Ed Page (5) Prof. Vinayak D. Paricharak M.Sc., B.Ed
 DNYANPRASAD / BRILLIANT'S ACADEMY
MHT-CET-XII - New Syllabus (MH) 2024-25
Time : 45 Min Chem : Full Portion Paper Marks : 50
Hints and Solutions

51) Ans: D) q+W

Sol: q + W = U, U is a state function 58) Ans: C) −119 kJmol−1
Therefore, q+W is a state function.

52) Ans: B) nucleophilic substitution.

53) Ans: C) 1
1 2 Sol:
Sol: Bond order in O22- = (10 − 8) = =1
2 2 rH = (B.E.)C=C + B.E.(H−H) + 4 (B.E.)C−H 
 
54) Ans: D) t99% = 2  t90% − ( B.E.)C−C + 6 ( B.E.)C−H 
Sol: For 99% completion of the reaction
r H = ( B.E.)C=C + B.E.(H−H) − (B.E.)(C−C) − 2(B.E.)C−H 
If a = 100; (a - x) = (100 - 99) = 1  
2.303 100 2.303 2 = 606 + 431 - 336 - 2 (410)
t99% = log = log k
k 1 k = −119 kJmol−1
2.303  2 4.606
= k t 99% = k ...(i)
For 90% completion of the reaction : 59) Ans: B) One
If a = 100; (a - x) = (100 - 90) = 10 Sol: Unimolecular layer.

2.303 100 60) Ans: B) Sugar and base

t90% = log ...(ii)
k 10 Sol: Sugar and base are present in nucleoside.
On dividing equation (i) by equation (ii),
t99% 4.603 k 61) Ans: A) Oleic acid
=  =2 Sol: Except oleic acid stearic acid, lauric acid as
t k 2.303
90%
well as palmitic acid are saturated fatty acid. Oleic
It means that time required for 99% completion of acid is an unsaturated fatty acid.
the reaction is twice the time required to complete
90% of the reaction.
62) Ans: C) Li+  Na+  K+  Rb+  Cs+
Sol: Smaller the cation, greater is the degree of its
55) Ans: D) All of the above
hydration.
Sol: The application of nanotechnology to food
As the degree of hydration decreases from
science and technology are agriculture, food safety
and biosecurity, product development, food Li+ to Cs+ , the hydration energy of alkali metals
processing and ingredient technologies. also decreases from Li to Cs+ .
Therefore the sequence in
56) Ans: D) 3
Li+  Na+  K+  Rb+  Cs+
Sol: CO2+ AR3d7 4S0 :
63) Ans: D) 65.31%
Atomic mass of oxygen
Sol: % composition = 100
3 unpaired electrons. Molar mass
64
= 100 = 65.3%
57) Ans: C) AuCu3 98
Sol: One eight of each corner atom (Au) and one
half of each face centred atom (Cu) are contained 64) Ans: D) HClO  HClO2  HClO3  HClO4
within the unit cell of the compound. Sol: On increasing oxidation number, the acidic
So, the number of Au atoms per unit cell strength also increases.
1
= 8  = 1 and the number of Cu atoms per unit
8 65) Ans: A) methyl iodide and phenol.
1
cell = 6  = 3 .
2
The formula of the compound is AuCu3 .

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 DNYANPRASAD / BRILLIANT'S ACADEMY

Sol: Rate of reaction is quadrupled on doubling the

concentration. Therefore r  [A]2 i.e. n = 2.

79) Ans: C) bring down the specific heat of water.

Sol: Sol: Ethylene glycol is added for lowering down the
freezing point of water in order that it does not
66) Ans: A) 600s freeze.
0.693 693 10−3
Sol: t1/2 = = 80) Ans: A) PVC
0.093 693 10−4
= 10 min = 600s. 81) Ans: A) Separation between electrode and
solution of it's ions.
67) Ans: D) Sol: Separation between electrode and solution of
it's ions.
(CH3 )2 NH  CH3NH2  (CH3 )3 N  NH3
Sol: As alkyl groups are electron releasing, electron
density on the nitrogen atom is increased and 82) Ans: C) Linkage isomerism, ionisation
alkylamines are more basic than ammonia. isomerism and geometrical isomerism
As a result, the basic character should decrease in Sol: Ionisation isomerism arises when the
the order, 3o amine  2o amine  1o amine. But coordination compounds give different ions in
because of the combined effect i.e., inductive effect solution. +
(+ I), steric factors and solvation of ions the basicity Co (NH ) (NO )  Cl Co (NH ) (NO )  + Cl−
decreases in the order, 2o amine  1o amine  3o  3 4 2 2  3 4 2 2

amine  NH3 for lower aliphatic amines. Co ( NH3 ) ( NO2 ) Cl  NO2
 4  +
Co (NH ) (NO ) Cl + NO−
68) Ans: C) +6  3 4 2  2
Sol: Due to absence of vacant d-orbitals, oxygen Linkage isomerism occurs in complex compounds
can not show +6 oxidation state. which contain ambidentate ligands like
NO2−, SCN− and CN− .
69) Ans: A) methyl chloride
Sol: CH3CH2 − COOAg + CH3Cl → Co (NH3 ) (NO2 )  Cl and Co (NH3 ) (ONO)  Cl
4 2 4 2
   
CH3 − CH2 − COOCH3 + AgCl −
are linkage isomers as NO 2 is linked through N or
through O. Octahedral complexes of the type
70) Ans: D) Pentose Ma4b2 exhibit geometrical isomerism.

71) Ans: C) C9H20

72) Ans: D) 6
Sol: Fe2+ has an electronic configuration 3d6 4s0 .

73) Ans: B) PCl5(g) → PCl3(g) + Cl2(g)

Sol: PCl5(g) → PCl3(g) + Cl2(g)
83) Ans: D) both (a) and (b)
74) Ans: A) 3-Methylphenol
84) Ans: B) one co-ordinate site.
75) Ans: A) Titanium
Sol: Ziegler-Natta catalyst is (CH3CH2 )3 Al + TiCl4 . 85) Ans: B) substitution
Sol: H-atoms of benzene are substituted by
chlorine
76) Ans: C) Above which the gas cannot be
liquefied how so ever high pressure may be applied 86) Ans: D) Optical activity
Sol: The critical temperature is the temperature
above which the gas cannot be liquefied how so 87) Ans: A) isopropyl alcohol.
ever high pressure may be applied.

77) Ans: C) +3 88) Ans: A) POCl3, H3PO4

Sol: In K2C2O4, Oxidation State of K=+1, Oxidation Sol: PCl5 + H2O → POCl3 + 2HCl
State of O=-2 X

 Oxidation State of carbon=2(+1)+2(x)+4(-2)=O POCl3 + 3H2O → H3PO4 + 3HCl

Y
 2x=8-2=6  x=+3
89) Ans: C) H2 gas at the cathode.
78) Ans: C) 2

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 DNYANPRASAD / BRILLIANT'S ACADEMY

Sol: 2H++ 2e−→ H2(g) , so H2 gas at cathode. 99) Ans: B) 6s2, 4f1−14,5d0−1
Sol: 6s2, 4f1−14,5d0−1

90) Ans: B) 100) Ans: A) The pH of 110−8M HCl is 8.

Sol: As pH = 8 is basic nature but HCl is a strong
acid.

Sol:
10 carbon-A carbon atom attached to one carbon
atom only is termed as 10 carbon or primary
carbon.
20 carbon - A carbon atom attached to two carbon
atoms is called 20 carbon or secondary carbon.
30 carbon -A carbon atom attached to three
carbon atoms is called 30 carbon or tertiary
carbon.
40 carbon -A carbon atom attached to four carbon
atoms is called 40 carbon or quaternary carbon.

91) Ans: D) Ziegler Natta catalyst.

Sol: Al(C2H5 )3 + TiCl4 represents Ziegler Natta
catalyst.

92) Ans: B) saponification.

6.02 1023
Ans: D)
93) 4
Sol: 6.023 1023 atoms are present in gm of
helium.
6.023 1023
1 gm of helium contains
4

94) Ans: A) 1m
Sol: Tb = Kb  molality(m)
if molality = 1 molal then
Tb = Kb

95) Ans: C) 8

96) Ans: A) weak acid and conjugate base.

97) Ans: A)
Formula weight of the desired product
 100
Sumof formula weight of all the reactants
Sol:
Formula weight of the desired product
 100
Sumof formula weight of all the reactants

98) Ans: C) R − CO − R'

Sol:

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