Tips and Tricks of The Day


Record Limit : 

    2019-Oct-18
Ex: A compound X with seven carbon atoms on treatment with Br2 and KOH gives Y. Y gives carbylamine test and upon diazotisation and coupling with phenol gives azodye. X is
(A) C6H5CONH2
(B) CH­3 – (CH2)5 – CONH2
(C) CH3 –C(CH3)2 - CH­2 – CH2 – CONH2
(D) O – CH3 – C6H4NH2
Explanation: Since Y gives coupling reaction after diazotisation it suggest that Y can be aniline or benzener ring substituted aniline. Since Y has been obtained from Hoffmann bromamide it means has – CO NH2 group with benzene ring. Hence it is C6H5CONH2
    2019-Oct-17
Ex: A narrow electron beam passes undeviated through an electric field E = 3×104 volt/mand an overlapping magnetic field B = 2×10−3 Weber/m2. If electric field and magnetic field are mutually perpendicular. The speed of the electrons is 
A) 60 m/s                                    
B) 10.3×107 m/s
C) 1.5×107 m/s  
D) 0.67×10−7 m/s
Explanation:  eE = evB ⇒ v = E/B = 3×104 / 2×10−3 = 1.5×107m/s
    2019-Oct-16
Ex: Two droplets merge with each other and forms a large droplet. In this process
A) Energy is liberated      
B) Energy is absorbed
C) Neither liberated nor absorbed
D) Some mass is converted into energy
Explanation: When two droplets merge with each other, their surface energy decreases. W = T(ΔA) = (negative) i.e. energy is released.
    2019-Oct-15
Ex: Formation of ammonia from H2 and N2 by Haber's process using Fe is an example of
A) Heterogeneous catalysis                             
B) Homogeneous catalysis
C) Enzyme catalysis                             
D) Non-catalytic process
Explanation: The catalytic process in which the reactants and the catalyst are in different phases is known as heterogenous catalysis.
                     Fe(S)
N2(g)+3H2(g) ---→  2NH3(g) The reactants are in gaseous state while the catalyst is in solid state.
    2019-Oct-11
Ex: The oxidation state of cobalt in the complex compound [Co(NH3)6]Cl3 is
A) + 3
B) + 6
C) + 5
D) + 2
Explanation: [Co(NH3)6]Cl3 → [Co(NH3)6]3+ +3Cl x + 6(0) = +3 ⇒ x = +3.
    2019-Oct-10
Ex: The statement that is true for the long form of the periodic table is
A) It reflects the sequence of filling the electrons in the order of sub-energy levels s, p, d and f
B) It helps to predict the stable valency states of the elements
C) It reflects trends in physical and chemical properties of the elements
D) It helps to predict the relative ionicity of the bonds between any two elements
Explanation: It reflects trends in physical and chemical properties of the elements.
    2019-Oct-09
Ex: The critical temperature of water is higher than that of O2 because H2O molecule has
A) Fewer electrons than oxygen
B) Two covalent bonds
C) V-shape
D) Dipole moment
Explanation: Critical temperature of water is more than O2 due to its dipole moment (Dipole moment of water = 1.84 D; Dipole moment of O2 = zero D).
    2019-Oct-07
Ex: An electromagnetic wave, going through vacuum is described by E = E0sin(kx−ωt). Which of the following is independent of wavelength
A) k    
B) w
C) k/w                                          
D) kw
Explanation: The angular wave number k = 2π/λ where l is the wave length. The angular frequency is w = 2πν. The ratio k/ω = 2π / λ 2πν = 1/νπ = 1 / c = constant.
    2019-Oct-04
Ex: When a potential difference is applied across, the current passing through?
A) An insulator at 0K is zero
B) A semiconductor at 0K is zero
C) A metal at 0K is finite
D) A P-N diode at 300K is finite, if it is reverse biased
Explanation: At 0 K, a semiconductor becomes a perfect insulator. Therefore at 0 K, if some potential difference is applied across an insulator or a semiconductor, current is zero. But a conductor will become a superconductor at 0 K. Therefore, current will be infinite. In reverse biasing at 300 K through a P-N junction diode, a small finite current flows due to minority charge carriers.
    2019-Oct-03
Ex: A linear harmonic oscillator of force constant 2×106N/m times and amplitude 0.01 m has a total mechanical energy of 160 joules. Its
A) Maximum potential energy is 100 J
B) Maximum K.E. is 100 J      
C) Maximum P.E. is 160 J
D) Minimum P.E. is zero
Explanation: Harmonic oscillator has some initial elastic potential energy and amplitude of harmonic variation of energy is  1/2 Ka2 =    1/2 × 2 × 106(0.01)2 = 100J | energy of the oscillator. Thus Kmax = 100J