Question

ꜱᴩᴇᴇᴅ ᴏꜰ ꜰᴀʟʟɪɴɢ ᴏʙᴊᴇᴄᴛ ꜰʀᴏᴍ ʜᴇɪɢʜᴛ ʜ ᴅᴇᴩᴇɴᴅꜱ ᴜᴩᴏɴ ɢʀᴀᴠɪᴛᴀᴛɪᴏɴᴀʟ ꜰᴏʀᴄᴇ ᴀɴᴅ ʜᴇɪɢʜᴛ ʜ ʙᴏᴛʜ.ᴜꜱɪɴɢ ᴅɪᴍᴇɴꜱɪᴏɴᴀʟ ᴀɴᴀʟyꜱɪꜱ ꜱʜᴏᴡ ᴛʜᴀᴛ ᴛʜɪꜱ ꜱᴩᴇᴇᴅ ɪꜱ ɢɪᴠᴇɴ ʙy ᴠ=ᴋ√ɢʜ​

Answer 1

Answer:

Answer refers to the attachment...hope it helps you .

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Answer 2

To proove :'

$v = k \sqrt{gh}$

Dimensions :'

$V = LT {}^{ - 1}$

$G = LT {}^{ - 2}$

$H = L$

From Dimension Rule

$V= KG {}^{x} H {}^{y}$

or

${LT {}^{ - 1} } = K(LT {}^{ - 2} {}^{x} )(L) {}^{y}$

$LT {}^{ - 1} = k(L) {}^{x + y} (T) {}^{ - 2x}$

Comparing both sides :'

$x + y = 1$

$- 2x = - 1$

$x = \frac{1}{2}$

$y = \frac{1}{2}$

$V= KG {}^{ \frac{1}{2} } H {}^{ \frac{1}{2} }$

$= K \sqrt{} GH$

${\pink{\tt{Hence-proved}}}$

$\sf\red{hope\:it\:helps\:you}$