SD STUDY FOR FUN

A body cools in 10 minutes from 60°C to 40°C. What will be its temperature after 10 minutes ? The temperature of the surroundings is 10°C. According to Newton's law of cooling, Solving equation (i) and (i), we get T=28°C

Q: If a, b, c  ∈  R then find a relation between a, b, c so that two quadratic equations a x 2 +bx+c =0 and 1003 x 2  1505x + 2007 = 0 have a common root Ans: We observe that the discriminant of the quadratic equation 1003 x 2 + 1505x + 2007 =0 is (1505)² - 4 x 1003 x 2007 < 0. ⇒ Roots of the equation are complex We know that complex roots of an equation occur in conjugate pairs, and in this case whenever  α  be a common root between given equations, other root automatically becomes common Hence the given equations have both roots common and consequently which is the required relation

The measure of length and breadth of a rectangle are l = (30.0 ± 0.2) cm and b = (10.0 ± 0.1) cm. What is the percentage error in the calculation of area of rectangle? Length =  30.0   ± 0.2 cm Breadth = 10.0 ± 0.1 cm Area  = Length  ×  Breadth   = (30.0 × 10.0) = 19.38 cm² ΔA = A × (ΔL/L + ΔB/B) = 300.00 × (0.2/30.0 + 0.1/10.0) = 5 cm² Area of rectangle = (300.00 ± 5.0) cm² percentage error in the calculation of area of rectangle =  ΔA/A  × 100  = 5/300  × 100  = 1.67%

Q. A diwali cracker of mass 60 g, at rest, explodes into three pieces A, B and C of mass 10 g, 20 and 30 g respectively. After explosion velocities of A and B are 30 m/s along east and 20 m/s along north respectively. Find the instantaneous velocity of C. Ans:  

Q:  In a battle, 70% of the combatants lost one eye, 80% lost one leg, 85% lost ear, 90% an arm. If x% lost all the four organs, then minimum value of x is Ans: Let, 7 0% of the combatants who lost one eye = A 80%  of the combatants who   lost one leg = B 85%  of the combatants who   lost ear = C 90%  of the combatants who   an arm = D Then, The combatants who lost one eye and one leg n(A ∩ B) =  n(A) + n(B) - n(AUB) ⇒ n(A ∩ B) = 70% +80% - 100% ⇒ n(A ∩ B) 50% Similarly we can find out as follows: The combatants who lost one leg and one ear = 65% The combatants who lost one ear and one arm = 75% The combatants who lost one arm and one eye = 60% Using n(AUBUCUD) =  n(A) + n(B) + n(C) + n(D) -n(A ∩ B) -n(B ∩ C) - n(C ∩ D) - n(A ∩ D) + n(A ∩ B ∩ C ∩ D) ⇒ n(A ∩ B ∩ C ∩ D) =  n(AUBUCUD) - n(A) -  n(B) - n(C) - n(D) + n(A ∩ B) + n(B ∩ C) + n(CND) + n(A ∩ D) ⇒ n(A ∩ B ∩ C ∩ D ) = 100% - 70 % - 80% - 85% - 90% + 5...

If SecA = x + 1/4x , Then prove that SecA + TanA = 2x or 1/2x SecA = x  +1/4x ∴, Sec²A = (x+1/4x)² = x² + 2.x.1/4x + 1/16x² = x² + 1/2 + 1/16x² Now,     Sec²A - Tan²A = 1 ⇒   Tan²A  = Sec²A - 1 ⇒   T an²A = x² + 1/2 + 1/16x² - 1 ⇒   T an²A = x² + 1/16x² - 1/2 ⇒   T an²A = x² - 2.x.1/4x + 1/16x² ⇒   T an²A = (x-1/4x)² ⇒   T anA   =  ±   (x-1/4x) ∴ CASE 1:    T anA = x - 1/4x S ecA +  T anA  = x + 1/4x + x -1/4x = 2x CASE  2:   T anA = -(x - 1/4x) SecA +  T anA = x + 1/4x -  ( x - 1/4x ) = x + 1/4x - x +  1/4x = 1/4x + 1/4x = 2/4x = 1/2x (Proved)