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A water tank stands on the roof of a building as shown. Then the value of 'h' for which the distance covered by the water 'x'  is greatest is - 
  • a)
    0.5 m
  • b)
    0.67 m
  • c)
    1 m 
  • d)
    none of these
Correct answer is option 'C'. Can you explain this answer?

Rohit Jain answered
 

the roots of x are (0,4) and the maximum of x is at h = 2.
The permitted value of h is 0 to 1 clearly h = 1 will give the maximum value of x is this interval. 
Aliter : If the column of water itself were from ground upto a height of 4m, h = 2m would give the maximum range x. Farther the hole is from this midpoint ,lower the range.Here the nearest point possible to this midpoint is the base of the container. Hence h = 1m.

Product of 22 integers is equal to 1, then their sum cannot be
  • a)
    0
  • b)
    1
  • c)
    2
  • d)
    4
Correct answer is option 'A,B,D'. Can you explain this answer?

Only integers which can satisfy this condition are -1 & +1. taking 10 times -1 and 12times +1 we will get the product 1. and sum will result in 2. check other cases similarly.

Figure shows a uniformly charged hemispherical shell. The direction of electric field at point p, that is off-centre (but in the plane of the largest circle of the hemisphere), will be along     
 
  • a)
    pa
  • b)
    pb
  • c)
    pc
  • d)
    pd
Correct answer is option 'B'. Can you explain this answer?

Let electric field at point. 'p' has both x and y component.
So similar electric field will be, for other hemisphere (upper half).
Now lets overlap both.

(Enet)p = 2 Ex and it should be zero (as E inside a full shell = 0).
So Ex = 0, So electric field at 'p' is purely in y direction.

Which of the following statements is/are correct ?      
  • a)
    Out of trimethylamine and trimethylphosphine, trimethylamine has higher dipole moment ?
  • b)
    Out of (SiH3)2O and (CH3)2O, (SiH3)2O is more basic. 
  • c)
    C–C bond length (in pm) in C2 molecule is greater than O–O bond length in O2 molecule.
  • d)
    N(SiMe3)3 and BF3 molecules are isostructural.
Correct answer is option 'A,C,D'. Can you explain this answer?

Lavanya Menon answered
(A) Nitrogen is more electronegative than phosphorus.
So, dipole moment of trimethylamine is greater than trimethy phosphine.
In trisilyl ether the lone pair of electron on oxygen atom is less easily available for donation because of pπ-dπ
delocalisation due to presence of the vacant d-orbital with Si. This however is not possible with carbon in CH3–O–CH3 due to the absence of d-orbital making it more basic.
(C) Bond order of C2
and O2 are same i.e., 2. In C2 molecules both bonds are π-bonds whereas, there is one  σ and one π-bond in O2 molecule
C2 = 131 pm ; O2 = 121 pm.

Match the statements in column-I with the statements in column-II.
            Column-I                                                            Column-II 
  (A)  A tight string is fixed at both ends and         (p)   At the middle, antinode is formed 
        sustaining standing wave                                       in odd harmonic
  (B)  A tight string is fixed at one end and           (q)   At the middle, node is formed 
        free at the other end                                              in even harmonic
  (C)  Standing wave is formed in an open organ   (r)    At the middle, neither node nor 
        pipe. End correction is not negligible.                     antinode is formed
  (D)  Standing wave is formed in a closed           (s)    Phase difference between SHMs of any
        organ pipe. End correction is not negligible.            two particles will be either p or zero.  
                                                                        (t)    The displacement of the particle in the                                                                                                                                                        middle is always non zero.
  • a)
    (A) p,q,s   (B) r,s  (C) s  (D) r,s 
  • b)
    (A) r,q   (B) s  (C) p,q,r  (D) p 
  • c)
    (A) r   (B) p,q  (C) r,p  (D) p,q 
  • d)
    (A) r  (B) q  (C) p,q  (D) r 
Correct answer is option 'A'. Can you explain this answer?

Aadhar Academy answered
(A) Number of loops (of length λ/2) will be even or odd and node or antinode will respectively be formed at the
middle.
Phase of difference between two particle in same loop will be zero and that between two particles in adjacent
loops will be π.
(B) and (D) Number of loops will not be integral. Hence neither a node nor an antinode will be formed in in the
middle.
Phase of difference between two particle in same loop will be zero and that between two particles in adjacent
loops will be π.
(C) Number of loops (of length λ/2) will be even or odd and antinode or node will respectively be formed at the
middle.
Phase of difference between two particle in same loop will be zero and that between two particles in adjacent loops will be π ..

In the shown circuit involving a resistor of resistance R W, capacitor of capacitance C farad and an ideal cell of emf E volts, the capacitor is initially uncharged and the key is in position 1. At t = 0 second the key is pushed to position 2 for t0 = RC seconds and then key is pushed back to position 1 for t0 = RC seconds. This process is repeated again and again. Assume the time taken to push key from position 1 to 2 and vice versa to be negligible.
Q. The charge on capacitor at t = 2RC second is 
  • a)
    CE
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

For t = 0 to t0 = RC seconds, the circuit is of charging type. The charging equation for this time is

Therefore the charge on capacitor at time t0 = RC is q0 
For t = RC to t = 2RC seconds, the circuit is of discharging type. The charge and current equation for this time are
 and  
Hence charge at t = 2 RC and current at t = 1.5 RC are

and 
Since the capacitor gets more charged up from t = 2RC to t = 3RC than in the interval  t = 0 to t = RC, the graph representing the charge variation is as shown in figure 

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