Tips and Tricks of The Day

Record Limit : 

Ex: A lorry and a car moving with the same K.E. are brought to rest by applying the same retarding force, then
Stopping distance = kinetic energy / retarding force
s = 1/2 mu2/F
If lorry and car both possess same kinetic energy and retarding force is also equal then both come to rest in the same distance.
Ex: The dimensions of an atom are of the order of an Angstrom. Thus there must be large electric fields between the protons and electrons. Why, then is the electrostatic field inside a conductor zero?
Explanation: Electrostatic fields are caused by excess charges. However, there is no excess charge on the inner surface of an isolated conductor. Therefore, electrostatic field inside a conductor is zero.
Ex: Which of the following is not a water absorber and dehydrating substance:
A) Silica gel
B) P2O5
C) Conc. H2SO4
D) Aqueous CaCl2
Explanation: Aqueous CaCl2 or hydrated CaCl2 cannot act as dehydrating agent.
Ex: The number and type of bonds between 2 carbon atoms in N2.
A) One sigma (σ) and one pi(π) bond
B) One sigma O2 and two pi (π) bond
C) One sigma (σ) and half pi (π) bond
D) One sigma CO2 bond
Explanation: CaC2 have one sigma and two p bond.
Ex: Find the atomic number of element belonging to 4th period and 17th group in Modem periodic table:
A) 17
B) 25           
C) 59
D) 35
Explanation: Element is Bromine (Z=35)
Ex: Calculate the work done, if a wire is loaded by 'Mg' weight and the increase in length is 'l' 
Work done = 1/2 Fl = Mgl / 2
Ex: A string on a musical instrument is 50 cm long and its fundamental frequency is 270 Hz. If the desired frequency of 1000 Hz is to be produced, the required length of the string is:
A) 13.5 cm                                   
B) 2.7 cm
C) 5.4 cm                                     
D) 10.3 cm
n ∝ 1/l ⇒ l2 / l1 = n1/n2 ⇒ l2 = l1(n1/n2) =50 × 270 / 1000 = 13.5 cm
Ex: For a reaction, the rate constant is expressed ask = Ae−40000/T. The energy of the activation is
A) 40000 cal        
B) 88000 cal
C) 80000 cal        
D) 8000 cal
k = Ae−Ea/RT
∴ - Ea/R = − 40000 
∴ Ea = 40000×2 = 80000 cal
Ex: Ethyne on reaction with water in the presence of HgSO4 and H2SO4 gives
A) Acetone              
B) Acetaldehyde
C) Acetic acid          
D) Ethyl alcohol                    HgSO4/H2SO4
Explanation: HC ≡ CH + H2O  --------------> CH3CHO
Ex: On complete combustion 1.4 g hydrocarbon gave 1.8 g water. Empirical formula of the hydrocarbon is
B) CH2
C) CH3
D) CH4
1.8gm water obtained from 1.4gm hydrocarbon 
∴18gm water obtained from - 1.4 / 1.8 × 18 = 14 gm.     
Empirical formula Mass = 14 ∴ Empirical formula = CH2