{"id":640228,"date":"2023-06-26T07:58:39","date_gmt":"2023-06-26T02:28:39","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/question\/in-the-given-figure-from-a-cuboidal-solid-metallic-block-of-dimensions-15-cmx10-cmx5-cm-a-cylindrical-hole-of-diameter-7-cm-is-drilled-out-find-the-surfac\/"},"modified":"2025-06-26T14:43:21","modified_gmt":"2025-06-26T09:13:21","slug":"in-the-given-figure-from-a-cuboidal-solid-metallic-block-of-dimensions-15-cmx10-cmx5-cm-a-cylindrical-hole-of-diameter-7-cm-is-drilled-out-find-the-surfac","status":"publish","type":"questions","link":"https:\/\/infinitylearn.com\/surge\/question\/mathematics\/in-the-given-figure-from-a-cuboidal-solid-metallic-block-of\/","title":{"rendered":"In the given figure, from a cuboidal solid metallic block of dimensions 15\u00a0cm\u00d710\u00a0cm\u00d75\u00a0cm, \u00a0 a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block. [Use \u03c0=227]"},"content":{"rendered":"","protected":false},"author":1,"template":"","meta":{"_yoast_wpseo_focuskw":"","_yoast_wpseo_title":"","_yoast_wpseo_metadesc":"In the given figure, from a cuboidal solid metallic block of dimensions 15\u00a0cm\u00d710\u00a0cm\u00d75\u00a0cm, \u00a0 a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block. [Use \u03c0=227]","custom_permalink":"question\/mathematics\/in-the-given-figure-from-a-cuboidal-solid-metallic-block-of\/"},"categories":[],"acf":{"question":"

In the given figure, from a cuboidal solid metallic block of dimensions <\/span><\/p>

\r\n\r\n \r\n \r\n 15<\/m:mn> <\/m:mo>cm<\/m:mtext>\u00d7<\/m:mo>10<\/m:mn> <\/m:mo>cm<\/m:mtext>\u00d7<\/m:mo>5<\/m:mn> <\/m:mo>cm<\/m:mtext>,<\/m:mo><\/m:mrow>\r\n <\/m:semantics><\/math>  a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block. [<\/span>Use \u03c0<\/mi>=<\/mo>22<\/mn><\/mrow>7<\/mn><\/mrow><\/mfrac>]<\/mo><\/math><\/p>

\"\"<\/p>

<\/p>

<\/p><\/spanstyle=\"-aw-import:ignore\"><\/pstyle=\"margin-top:0pt;margin-bottom:0pt;line-height:normal\"><\/spanstyle=\"-aw-import:ignore\"><\/pstyle=\"margin-top:0pt;margin-bottom:0pt;line-height:normal\">","options":[{"option":"550 cm2<\/span>","correct":false},{"option":"583 cm2<\/span>","correct":true},{"option":"500 cm2<\/span>","correct":false},{"option":"400 cm2<\/span> <\/span>","correct":false}],"solution":"Given that a cuboidal solid metallic block has dimensions <\/span>\r\n\r\n \r\n \r\n 15<\/m:mn> <\/m:mo>cm<\/m:mtext>\u00d7<\/m:mo>10<\/m:mn> <\/m:mo>cm<\/m:mtext>\u00d7<\/m:mo>5<\/m:mn> <\/m:mo>cm<\/m:mtext><\/m:mrow>\r\n <\/m:semantics><\/math>  <\/span>and a cylindrical hole of diameter 7 cm is drilled out.<\/span>
\"\"Let l, b and h be the length, breadth and height of the cuboid respectively and r and h be the radius and height of the cylinder respectively.<\/span>
Here, <\/span>l<\/mi>=<\/mo>15<\/mn>cm<\/mi>,<\/mo> b<\/mi>=<\/mo>10<\/mn>cm<\/mi>,<\/mo> h<\/mi>=<\/mo>5<\/mn> cm<\/mi>,<\/mo> r<\/mi>=<\/mo>7<\/mn><\/mrow>2<\/mn><\/mrow><\/mfrac>cm<\/mi>.<\/mo><\/math>
The surface area of the cuboid is given by,<\/span>
\r\n\r\n \r\n \r\n S<\/m:mi>A<\/m:mi>=<\/m:mo>2<\/m:mn>\r\n \r\n l<\/m:mi>b<\/m:mi>+<\/m:mo>b<\/m:mi>h<\/m:mi>+<\/m:mo>h<\/m:mi>l<\/m:mi><\/m:mrow>\r\n <\/m:mfenced><\/m:mrow>\r\n <\/m:semantics><\/math>  <\/span>\r\n\r\n \r\n \r\n \r\n \r\n \u21d2<\/m:mo>S<\/m:mi>A<\/m:mi>=<\/m:mo>2<\/m:mn>\r\n \r\n 15<\/m:mn>\u00d7<\/m:mo>10<\/m:mn>+<\/m:mo>10<\/m:mn>\u00d7<\/m:mo>5<\/m:mn>+<\/m:mo>15<\/m:mn>\u00d7<\/m:mo>5<\/m:mn><\/m:mrow>\r\n <\/m:mfenced>\r\n <\/m:mtd>\r\n <\/m:mtr>\r\n \r\n \r\n \u21d2<\/m:mo>S<\/m:mi>A<\/m:mi>=<\/m:mo>2<\/m:mn>(<\/m:mo>150<\/m:mn>+<\/m:mo>50<\/m:mn>+<\/m:mo>75<\/m:mn>)<\/m:mo>\r\n <\/m:mtd>\r\n <\/m:mtr>\r\n \r\n \r\n \u21d2<\/m:mo>S<\/m:mi>A<\/m:mi>=<\/m:mo>2<\/m:mn>(<\/m:mo>275<\/m:mn>)<\/m:mo>\r\n <\/m:mtd>\r\n <\/m:mtr>\r\n \r\n \r\n \u21d2<\/m:mo>S<\/m:mi>A<\/m:mi>=<\/m:mo>550<\/m:mn>c<\/m:mi>\r\n m<\/m:mi>\r\n 2<\/m:mn>\r\n <\/m:msup>\r\n \r\n <\/m:mtd>\r\n <\/m:mtr>\r\n <\/m:mtable>\r\n \r\n <\/m:semantics><\/math> 
The curved surface area of the cylinder is given by,<\/span>
 <\/span>\r\n\r\n \r\n \r\n C<\/m:mi>S<\/m:mi>A<\/m:mi>=<\/m:mo><\/m:mrow>\r\n <\/m:semantics><\/math>  <\/span>2<\/mn>\u03c0rh<\/mi><\/math> <\/span> <\/span>\r\n\r\n \r\n \r\n \r\n \r\n \u21d2<\/m:mo>C<\/m:mi>S<\/m:mi>A<\/m:mi>=<\/m:mo>2<\/m:mn>\u00d7<\/m:mo>\r\n \r\n 22<\/m:mn><\/m:mrow>\r\n 7<\/m:mn>\r\n <\/m:mfrac>\r\n \u00d7<\/m:mo>\r\n 7<\/m:mn>\r\n 2<\/m:mn>\r\n <\/m:mfrac>\r\n \u00d7<\/m:mo>5<\/m:mn>\r\n <\/m:mtd>\r\n <\/m:mtr>\r\n \r\n \r\n \u21d2<\/m:mo>C<\/m:mi>S<\/m:mi>A<\/m:mi>=<\/m:mo>22<\/m:mn>\u00d7<\/m:mo>5<\/m:mn>\r\n <\/m:mtd>\r\n <\/m:mtr>\r\n \r\n \r\n \u21d2<\/m:mo>C<\/m:mi>S<\/m:mi>A<\/m:mi>=<\/m:mo>110<\/m:mn>c<\/m:mi>\r\n m<\/m:mi>\r\n 2<\/m:mn>\r\n <\/m:msup>\r\n \r\n <\/m:mtd>\r\n <\/m:mtr>\r\n <\/m:mtable>\r\n \r\n <\/m:semantics><\/math> 
The area of the two ends of the cylindrical holes is,<\/span>
\r\n\r\n \r\n \r\n A<\/m:mi>=<\/m:mo>2<\/m:mn>\r\n \r\n \u03c0<\/m:mi>\r\n r<\/m:mi>\r\n 2<\/m:mn>\r\n <\/m:msup>\r\n <\/m:mrow>\r\n <\/m:mfenced><\/m:mrow>\r\n <\/m:semantics><\/math>  <\/span>\r\n\r\n \r\n \r\n \r\n \r\n \u21d2<\/m:mo>A<\/m:mi>=<\/m:mo>2<\/m:mn>\r\n \r\n \r\n \r\n 22<\/m:mn><\/m:mrow>\r\n 7<\/m:mn>\r\n <\/m:mfrac>\r\n \u00d7<\/m:mo>\r\n 7<\/m:mn>\r\n 2<\/m:mn>\r\n <\/m:mfrac>\r\n \u00d7<\/m:mo>\r\n 7<\/m:mn>\r\n 2<\/m:mn>\r\n <\/m:mfrac>\r\n <\/m:mrow>\r\n <\/m:mfenced>\r\n <\/m:mtd>\r\n <\/m:mtr>\r\n \r\n \r\n \u21d2<\/m:mo>A<\/m:mi>=<\/m:mo>\r\n \r\n 22<\/m:mn>\u00d7<\/m:mo>7<\/m:mn><\/m:mrow>\r\n 2<\/m:mn>\r\n <\/m:mfrac>\r\n \r\n <\/m:mtd>\r\n <\/m:mtr>\r\n \r\n \r\n \u21d2<\/m:mo>A<\/m:mi>=<\/m:mo>11<\/m:mn>\u00d7<\/m:mo>7<\/m:mn>\r\n <\/m:mtd>\r\n <\/m:mtr>\r\n \r\n \r\n \u21d2<\/m:mo>A<\/m:mi>=<\/m:mo>77<\/m:mn>c<\/m:mi>\r\n m<\/m:mi>\r\n 2<\/m:mn>\r\n <\/m:msup>\r\n \r\n <\/m:mtd>\r\n <\/m:mtr>\r\n <\/m:mtable>\r\n \r\n <\/m:semantics><\/math> 
Here,<\/span>
Surface area of remaining block = Surface area of cuboid + curved surface area of cylinder \u2013 area of ends of two cylindrical holes <\/span>
\u21d2 Surface area = <\/span><\/span>(<\/mo>550<\/mn> <\/mi>+<\/mo> <\/mi>110<\/mn> <\/mi>-<\/mo> <\/mi>77<\/mn>)<\/mo>cm<\/mi><\/mrow>2<\/mn><\/mrow><\/msup><\/math>
\u21d2 Surface area = 583<\/span><\/span>cm<\/mi><\/mrow>2<\/mn><\/mrow><\/msup><\/math>
Hence the surface area of remaining solid is 583 <\/span>cm<\/mi><\/mrow>2<\/mn><\/mrow><\/msup><\/math>.<\/span>
Therefore, the correct answer is option (2).<\/span>
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