SAI'S SCIENCE AND MATHS CLASSES

A body of   weight 2 kg is suspended as shown Find the tension T1 in   the horizontal string (KMEE 2002) Since the system is in equilibrium the resultant must be zero Let T be tension on the other string then resolve the tension T vertically and horizontally as Tv and T H then Tv= T sin30=2kgwt----------------(1) T H = T cos30=T1--------------------------(2) • • • T =4 kgwt from (1) substituting T in (2) , T1=4 cos30=2√3 • • • Tension in the horizontal string is 2√3 kgwt Prepared by Sai’s Classes Poonkunnam Thrissur +919846088761

If two physical quantities x and y are related as y =k x n if the percentage error on x is P derive an expression to find the percentage error on y y =k x n ------------------------(1) Δy= k n x (n-1) Δx -----------(2) (by differentiating   equation 1 ) Divide equation (2) by (1) Δy /y =n Δx/ x ( Δy /y) ×100=n   ( Δx/ x) ×100 ie percentage error on y = n times percentage error on x = n P Note that absolute eror in 'y' is denoted by  Δy , relative error by  Δy /y and percentage error by  ( Δy /y) ×100 so the general rules are if i) y=a+b or a-b then absolute error in 'y' is the sum of the absolute errors of 'a' and 'b' ie  Δy= Δa+ Δb ii) y=a × b or a/b  then the relative error in 'y ' is the sum of the relative errors of 'a' and 'b' ie  Δy /y= Δa /a+ Δb /b so same is the percentage error in 'y' iii)  y =k x n then the  relative error   in 'y ' is  n times  relative  e...

Notes on this topic is available on the home page link The radius of a sphere is expressed as 110.4 ± 0.2 cm Find the percentage error in    a) radius   b) surface area    c) volume a) Percentage error = ( Δx/x )×100 • • • percentage error in radius= (0.2/110.4) x 100 = 0.18% b) Surface area = 4 πr 2 • • • percentage error in Surface area = 2 × Percentage error in radius=2×0.18= 0.36% ( derivation for the expression will be posting   using calculus soon) c) Volume = 4/3(πr 3 ) • • • percentage error in volume = 3 × Percentage error in radius=3×0.18= 0.54% Prepared by Sai’s Classes Poonkunnam Thrissur +919846088761

Projectiles A bullet is to be fired with a speed of 1000 m/sec to hit a target 200 m away on a horizontal level If g=10 m/sec 2 At what height ( h ) the gun should be hold to fire and hit the target? 200 m h For a projectile, time taken to cover the horizontal distance = time taken to cover the vertical height as free fall   ie time taken to cover 200 m = time for free falling h m   time =distance / speed ( for uniform motion in the horizontal direction)   t=200 ÷ 1000 =0.2 sec   by   Newtons equations of motion   h=ut+½gt 2   u=0 ( . . . free fall)   h =0+½ ×10 × 0.04= 0.2 m   ie the gun must hold at a height of 20 cm above the target Prepared by Sai’s Classes Poonkunnam Thrissur +919846088761