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NCERT Class X| Physics Solutio,, §, , 16 Units, alas 5 nber of significant figures is 4, as aj nc, ¢) at oe ae wacinel and trailing zero are significant, Let,, ap o00 the number of significant figures is 4 (reason jg Same, CE ee ean — > ag, (e) in Pash, ue number of significant figures is 4 (reason is Same (b), e WO ag, i) a Pbeae| the number of signifiant figures is 4 (reason is sam, : as in part 'a’)., . fan ea i : ickness of a rectangu|e 11. The length, breadth and thie } J gular sh... We, Me crite 4,234 m, 1.005 m and 2.01 cm respectively. Give the areg a mae ;, volume of the sheet to correct significant figures. The, if different values of a same quantity are given in different Units, then fir 2%, 2G of all convert them in same units without changing the Number of valu, significant figures. whz, Solution Given, length (/) =4.234 m Sol, breadth (b) = 1005 m : Ma, thickness (t) = 2.01 cm =0.0201 m, Area of sheet (A) =2(/xb+bxt+tx/), =2[(4.234 x 1.005) + (1005 x 0.0201) + (0.0201 x4.234)| =, =2 x4.3604739 AP, =8,7209478 m? P, As thickness has least number of significant figures 3, therefore Bu, rounding-off area up to three significant figures, we get, Area of sheet (A) = 8.72 m2 oe, Volume of sheet (V) =/xb xt in, =4234 1005 x 0.0201, = 00855289, Rounding-off up to three significant figures, we get, Volume of the sheet = 0.0855 m3, A, Question 12. The mass of a box measured by a grocer’s balance is P, , aoe Two gold pieces of masses 20,15 g and 20.17 are added to the box. XS, , Vhat is (a) the total mass of the box, (b) the difference in the mass of the C, Pleces to Correct significant figures? :, , ; fe, Solution Given, mass of the box (m) = 2.3 kg, Mass of first gold piece (m,) = 20,15 g, ae = 002015 k, Mass of Second gold piece (m) 20.17 g =002017 es, (a) Total mass of the box (M) = m + m +m :, 2, —-=234002015+002017 p, As the mass of th ra i