Advertisements
Advertisements
प्रश्न
Unequal masses will not balance on a fulcrum if they are at equal distance from it; one side will go up and the other side will go down.
Unequal masses will balance when the following proportion is true:
`("mass"1)/("length"2) = ("mass"2)/("length"1)`
Two children can be balanced on a seesaw when
`("mass"1)/("length"2) = ("mass"2)/("length"1)`. The child on the left and child on the right are balanced. What is the mass of the child on the right?
Advertisements
उत्तर
It is given that, for balancing.
`("Mass" 1)/("Length" 2) = ("Mass" 2)/("Length" 1)`
According to the question,
Mass 1 = 24 kg, length 1 = 3 m and length 2 = 2 m
∴ `24/2 = ("Mass" 2)/3` ...[By cross-multiplication]
⇒ Mass 2 = `(24 xx 3)/2` = 36 kg
APPEARS IN
संबंधित प्रश्न
Find the fourth proportion to the following:
3,5 and 15
Find the mean proportion of the following :
`28/3` and `175/27`
If ( a+c) : b = 5 : 1 and (bc + cd) : bd = 5 : 1, then prove that a : b = c : d
Find the third proportional to:
x - y, x2 - y2
Find the fourth proportional to:
x3 - y2, x4 + x2y2 + y4, x - y.
If q is the mean proportional between p and r, prove that
`p^2 - q^2 + r^2 = q^4[(1)/p^2 - (1)/q^2 + (1)/r^2]`.
If `a = (b + c)/(2), c = (a + b)/(2)` and b is mean proportional between a and c, prove that `(1)/a + (1)/c = (1)/b`.
Find the fourth proportional to 1.5, 2.5, 4.5
Show that the following numbers are in continued proportion:
16, 84, 441
Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion.
25 cm : 1 m and ₹ 40 : ₹ 160
